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```Notes

for an Introductory Course

On Electrical Machines

and Drives

E.G.Strangas

MSU Electrical Machines and Drives Laboratory

Contents

Preface ix

Three Phase Circuits and Power 1

1.1 Electric Power with steady state sinusoidal quantities 1

1.2 Solving 1 -phase problems 5

1.3 Three-phase Balanced Systems 6

1.4 Calculations in three-phase systems 9

15
15
15
19

21

22
25
27

29
29
30
32
36
37

2 Magnetics

2.1

Introduction

2.2

The Governing Equations

2.3

Saturation and Hysteresis

2.4

Permanent Magnets

2.5

2.6

Eddy Currents and Eddy Current Losses

2.7

Torque and Force

3 Transformers

3.1

Description

3.2

The Ideal Transformer

3.3

Equivalent Circuit

3.4

Losses and Ratings

3.5

Per-unit System

VI CONTENTS

3.6 Transformer tests 40

3.6.1 Open Circuit Test 41

3.6.2 Short Circuit Test 41

3. 7 Three-phase Transformers 43

3.8 Autotransformers 44

4 Concepts of Electrical Machines; DC motors 47
4.1 Geometry, Fields, Voltages, and Currents 47

5 Three-phase Windings 53

5.1 Current Space Vectors 53

5.2 Stator Windings and Resulting Flux Density 55
5.2.1 Balanced, Symmetric Three-phase Currents 58

5.3 Phasors and space vectors 58

5.4 Magnetizing current, Flux and Voltage 60

6 Induction Machines 63

6.1 Description 63

6.2 Concept of Operation 64

6.3 Torque Development 66

6.4 Operation of the Induction Machine near Synchronous Speed 67

6.5 Leakage Inductances and their Effects 71

6.6 Operating characteristics 72

6. 7 Starting of Induction Motors 75

6. 8 Multiple pole pairs 76

7 Synchronous Machines and Drives 81

7. 1 Design and Principle of Operation 81

7.1.1 Wound Rotor Carrying DC 81

7.1.2 Permanent Magnet Rotor 82

7.2 Equivalent Circuit 82

7.3 Operation of the Machine Connected to a Bus of Constant Voltage

and Frequency 84

7.4 Operation from a Source of Variable Frequency and Voltage 88

7. 5 Controllers for PMAC Machines 94

7.6 Brushless DC Machines 95

8 Line Controlled Rectifiers 99

8.1 1- and 3 -Phase circuits with diodes 99

8.2 One -Phase Full Wave Rectifier 100

8.3 Three-phase Diode Rectifiers 102

8.4 Controlled rectifiers with Thyristors 103

CONTENTS Vii

8.5 One phase Controlled Rectifiers 104
8. 5. 1 Inverter Mode 1 04

8.6 Three-Phase Controlled Converters 106

8.7 "Notes 107

9 Inverters 109

9.1 1 -phase Inverter 109

9.2 Three-phase Inverters 111

10 DC-DC Conversion 117

10.1 Step-Down or Buck Converters 117

10.2 Step-up or Boost Converter 119

10.3 Buck-boost Converter 122

Preface

The purpose of these notes is be used to introduce Electrical Engineering students to Electrical
Machines, Power Electronics and Electrical Drives. They are primarily to serve our students at
MSU: they come to the course on Energy Conversion and Power Electronics with a solid background
in Electric Circuits and Electromagnetics, and many want to acquire a basic working knowledge
of the material, but plan a career in a different area (venturing as far as computer or mechanical
engineering). Other students are interested in continuing in the study of electrical machines and
drives, power electronics or power systems, and plan to take further courses in the field.

Starting from basic concepts, the student is led to understand how force, torque, induced voltages
and currents are developed in an electrical machine. Then models of the machines are developed, in
terms of both simplified equations and of equivalent circuits, leading to the basic understanding of
modern machines and drives. Power electronics are introduced, at the device and systems level, and
electrical drives are discussed.

Equations are kept to a minimum, and in the examples only the basic equations are used to solve
simple problems.

These notes do not aim to cover completely the subjects of Energy Conversion and Power
Electronics, nor to be used as a reference, not even to be useful for an advanced course. They are
meant only to be an aid for the instructor who is working with intelligent and interested students,
who are taking their first (and perhaps their last) course on the subject. How successful this endeavor
has been will be tested in the class and in practice.

In the present form this text is to be used solely for the purposes of teaching the introductory
course on Energy Conversion and Power Electronics at MSU.

E.G.Strangas

E. Lansing, Michigan and Pyrgos, Tinos

IX

A Note on Symbols

Throughout this text an attempt has been made to use symbols in a consistent way. Hence a script
letter, say v denotes a scalar time varying quantity, in this case a voltage. Hence one can see

v = 5 sin cut or v = v sin cut

The same letter but capitalized denotes the rms value of the variable, assuming it is periodic.
Hence:

v = v2Vsinuit

The capital letter, but now bold, denotes a phasor:

V = Ve je

Finally, the script letter, bold, denotes a space vector, i.e. a time dependent vector resulting from
three time dependent scalars:

V = V\ + V2& 1 + v^ 21

In addition to voltages, currents, and other obvious symbols we have:
Magnetic flux Density (T)
Magnetic filed intensity (A/m)

Flux (Wb) (with the problem that a capital letter is used to show a time
dependent scalar)

flux linkages (of a coil, rms, space vector)

synchronous speed (in electrical degrees for machines with more than
two-poles)

rotor speed (in electrical degrees for machines with more than two-poles)
rotor speed (mechanical speed no matter how many poles)
angular frequency of the rotor currents and voltages (in electrical de-
grees)

Torque (Nm)
Real and Imaginary part of •

B

H

\$

A, A, A

UJ S

U)

u m

uo r

T

»(').

9f(-)

1

Three Phase Circuits and Power

Chapter Objectives

In this chapter you will learn the following:

• The concepts of power, (real reactive and apparent) and power factor

• The operation of three-phase systems and the characteristics of balanced loads in Y and in A

• How to solve problems for three-phase systems

1 .1 ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES

We start from the basic equation for the instantaneous electric power supplied to a load as shown in
figure 1.1

i(t)

+
v(t)

p(t) = i(t) ■ v(t)

(1.1)

1

2 THREE PHASE CIRCUITS AND POWER

where i(t) is the instantaneous value of current through the load and v(t) is the instantaneous value
of the voltage across it.

In quasi-steady state conditions, the current and voltage are both sinusoidal, with corresponding
amplitudes i and v 9 and initial phases, <pi and <p V9 and the same frequency, uo = 2tt/T — 2nf:

v(t) = vsm(ut + (j) v ) (1.2)

i(t) = ism(u;t + <pi) (1.3)

In this case the rms values of the voltage and current are:

V = J^£v[sin(u;t + (P v )} 2 dt=^= (1.4)

1 ? T

I = J-J i[sm(u;t + i )} z dt=-= (1.5)

and these two quantities can be described by phasors, V = V^ v and I = I L ^ 1 .
Instantaneous power becomes in this case:

p(t) = 2VI[sm(ut + <p v )sm(ujt + <pi)]

= 2VI- [cos((p v - (pi) + cos{2uot + (p v + (pi)) (1.6)

The first part in the right hand side of equation 1.6 is independent of time, while the second part
varies sinusoidally with twice the power frequency. The average power supplied to the load over
an integer time of periods is the first part, since the second one averages to zero. We define as real
power the first part:

P = VIcos((P v -&) (1.7)

If we spend a moment looking at this, we see that this power is not only proportional to the rms
voltage and current, but also to cos((p v — (pi). The cosine of this angle we define as displacement
factor, DF. At the same time, and in general terms (i.e. for periodic but not necessarily sinusoidal
currents) we define as power factor the ratio:

Pf=fj d.8)

and that becomes in our case (i.e. sinusoidal current and voltage):

pf = cos((p v -(pi) (1.9)

Note that this is not generally the case for non-sinusoidal quantities. Figures 1.2-1.5 show the cases
of power at different angles between voltage and current.

We call the power factor leading or lagging, depending on whether the current of the load leads
or lags the voltage across it. It is clear then that for an inductive/resistive load the power factor is
lagging, while for a capacitive/resistive load the power factor is leading. Also for a purely inductive
or capacitive load the power factor is 0, while for a resistive load it is 1 .

We define the product of the rms values of voltage and current at a load as apparent power, S:

S = VI (1.10)

ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES 3

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.2 Power at pf angle of 0°. The dashed line shows average power, in this case maximum

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.3 Power at pf angle of 30°. The dashed line shows average power

and as reactive power, Q

Q = VIsm(0 v -&)

(1.11)

Reactive power carries more significance than just a mathematical expression. It represents the
energy oscillating in and out of an inductor or a capacitor and a source for this energy must exist.
Since the energy oscillation in an inductor is 180° out of phase of the energy oscillating in a capacitor,

4 THREE PHASE CIRCUITS AND POWER

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.4 Power at pf angle of 90°. The dashed line shows average power, in this case zero

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Fig. 1.5 Power at pf angle of 180°. The dashed line shows average power, in this case negative, the opposite
of that in figure 1.2

the reactive power of the two have opposite signs by convention positive for an inductor, negative for
a capacitor.

The units for real power are, of course, W, for the apparent power V A and for the reactive power
VAr.

SOLVING 1 -PHASE PROBLEMS

Using phasors for the current and voltage allows us to define complex power S as:

and finally

For example, when

VI*

S = P + jQ

v(t)
i{t)

v/(2-120-sin(377t+^)l/
6

V^2-5-sin(377£+^)A

(1.12)
(1.13)

(1.14)

(1.15)
(1.16)

then S = VI = 120 • 5 = 600W, while pf = cos(tt/6 - tt/4) = 0.966 leading. Also:
S = VI* = 120 Z7r/6 5 Z_7r/4 = 579.6W - jlhh.WAr

(1.17)

Figure 1 .6 shows the phasors for lagging and leading power factors and the corresponding complex
power S.

► V

Fig. 1.6 (a) lagging and (b) leading power factor

1.2 SOLVING 1 -PHASE PROBLEMS

Based on the discussion earlier we can construct the table below:

Reactive

Capacitive
Resistive

Reactive power
Q>0
Q<0
Q =

Power factor

lagging

1

Q =

-- -sJi- P f 2 =

-QOkVAr ,

or

n) -

= sin [arccos0.8]

Q =

= S 8m((/) v - fa)

6 THREE PHASE CIRCUITS AND POWER

We also notice that if for a load we know any two of the four quantities, S, P, Q, pf, we can
calculate the other two, e.g. if S = lOOkVA, pf = 0.8 leading, then:

P = S-pf = 80kW

sin

Notice that here Q < 0, since the pf is leading, i.e. the load is capacitive.

Generally in a system with more than one loads (or sources) real and reactive power balance, but
not apparent power, i.e. P total = J2i P i> Q total =Y^iQi> but s total ^Y^i S i-

In the same case, if the load voltage were Vl = 2000 V\ the load current would be II = S/V
= 100 • 10 3 /2 • 10 3 = 50A If we use this voltage as reference, then:

V = 2000 Zo V
I = 50 Z ^ = 50 Z36 - 9 °A
S = V I* = 2000 Zo • 50 Z ~ 36 - 9 ° = P + jQ = 80 • 10 3 W - J60 • 10 3 V Ar

1 .3 THREE-PHASE BALANCED SYSTEMS

Compared to single phase systems, three-phase systems offer definite advantages: for the same power
and voltage there is less copper in the windings, and the total power absorbed remains constant rather
than oscillate around its average value.

Let us take now three sinusoidal-current sources that have the same amplitude and frequency, but
their phase angles differ by 120°. They are:

-f) (1.18)

If these three current sources are connected as shown in figure 1.7, the current returning though node
n is zero, since:

2tt 2tt

sm(uut + (j)) + sm(cut - (j) + — ) + sm(ut + <j> + — ) = (1.19)

h(t) --

= v27 sm(ut

i 2 (t) --

= v2I sm(ut

h(t) =

= v27 sm(ut

us also ta

ke three voltage sources:

v a (t) =

V2V sm(uot + (j))

v b (t) =

2tt

V2V sin(ujt + (f) )

o

Vc(t) =

2tt
V2V sin(ujt + (f) + — )
o

(1.20)

connected as shown in figure 1.8. If the three impedances at the load are equal, then it is easy
to prove that the current in the branch n — n' is zero as well. Here we have a first reason why

THREE-PHASE BALANCED SYSTEMS

Fig. 1.7 Zero neutral current in a ^-connected balanced system

Fig. 1.8 Zero neutral current in a voltage-fed, F-connected, balanced system.

three-phase systems are convenient to use. The three sources together supply three times the power
that one source supplies, but they use three wires, while the one source alone uses two. The wires
of the three-phase system and the one-phase source carry the same current, hence with a three-phase
system the transmitted power can be tripled, while the amount of wires is only increased by 50%.

The loads of the system as shown in figure 1 .9 are said to be in Y or star. If the loads are connected
as shown in figure 1.11, then they are said to be connected in Delta, A, or triangle. For somebody
who cannot see beyond the terminals of a Y or a A load, but can only measure currents and voltages
there, it is impossible to discern the type of connection of the load. We can therefore consider the
two systems equivalent, and we can easily transform one to the other without any effect outside the
load. Then the impedances of a Y and its equivalent A symmetric loads are related by:

1

Z A

(1.21)

Let us take now a balanced system connected in Y, as shown in figure 1.9. The voltages
between the neutral and each of the three phase terminals are Vi n = V^ , V 2n = V^^~^ , and
V 3n = V L ^ + "? . Then the voltage between phases 1 and 2 can be shown either through trigonometry
or vector geometry to be:

8 THREE PHASE CIRCUITS AND POWER

12

V

2n

Fig. 1.9 Y Connected Loads: Voltages and Currents

Fig. 1.10 Y Connected Loads: Voltage phasors

12

Vi - V 2 = V3V l ^i

(1.22)

This is shown in the phasor diagrams of figure 1.10, and it says that the rms value of the line-to-line
voltage at a Y load, Vu, is y/3 times that of the line-to-neutral or phase voltage, V\ n . It is obvious
that the phase current is equal to the line current in the Y connection. The power supplied to the
system is three times the power supplied to each phase, since the voltage and current amplitudes and
the phase differences between them are the same in all three phases. If the power factor in one phase
is pf = cos(0 v — <pi), then the total power to the system is:

P^<t> + jQ3(f>

3ViI*

VsVuIi cos(<

<t>i) + jy/SViiIi sin(

i)

(1.23)

Similarly, for a connection in A, the phase voltage is equal to the line voltage. On the other hand,
if the phase currents phasors are Ii 2 = I ^, I23 = I ^~~z and I31 = 7^^+^, then the current of

CALCULATIONS IN THREE-PHASE SYSTEMS 9

'31 - hi

"3

Fig. 1.11 A Connected Loads: Voltages and Currents

line 1, as shown in figure 1.11 is:

Ii=Ii2-l3i = V3/ Z ^-t
To calculate the power in the three-phase, Y connected load,

\$30 = ^3(1) + jQscj)

= 3ViI*

= y/SVuIi cos(0 v - 0i) + jVsVuIi sin(0 v

(1.24)

(1.25)

1 .4 CALCULATIONS IN THREE-PHASE SYSTEMS

It is often the case that calculations have to be made of quantities like currents, voltages, and power,
in a three-phase system. We can simplify these calculations if we follow the procedure below:

1. transform the A circuits to Y,

2. connect a neutral conductor,

3. solve one of the three 1 -phase systems,

4. convert the results back to the A systems.

1.4.1 Example

For the 3-phase system in figure 1.12 calculate the line-line voltage, real power and power factor at

First deal with only one phase as in the figure 1.13:

I = - ™ . r = 13.02 Z - 406 °^

Vin = IZ 1 = 13.02 Z - 40 - 6 °(7 + j5) = 111.97 Z - 5< V

S L ,i^ = V L I* = 1.186 -10 3 +j0.847- 10 3

P LH = 1.186fcW, Q L14> = 0.847W Ar

pf = cos(-5° - (-40.6°)) = 0.814 lagging

10 THREE PHASE CIRCUITS AND POWER

jifl

7+5jft

Fig. 1.12 A problem with Y connected load.

jin

120V

7+j5a

Fig. 1.13 One phase of the same load

For the three-phase system the load voltage (line-to-line), and real and reactive power are:

V L ,i-i = \/3- 111.97= 193.94V

Plm = 3.56kW, Q LM = 2MlkVAr

1.4.2 Example

For the system in figure 1.14, calculate the power factor and real power at the load, as well as the
phase voltage and current. The source voltage is 400 V line-line.

jin

18+j6£2

CALCULATIONS IN THREE-PHASE SYSTEMS 1 1

First we convert the load to Y and work with one phase. The line to neutral voltage of the source
is Vin = 400/^ = 231V.

Fig. 1. 15 The same load converted to Y

jlQ

231V

6+J2Q.

Fig. 1.16 One phase of the Y load

231 34.44 Z - 26 - 6 °A

J1 + 6+J2

V L = I L (6+j2) = 217.8 Z ~ 8 - 1 V

The power factor at the load is:

pf = cos(<^ - ((),) = cos(-8.1° + 26.6°) = 0M8lag

Converting back to A:

1\$ = Il/Vs = 34.44/ v^ = 19.88A
Vu = 217.8 -a/3 -377.22V

Ps<f> = VsVu I L pf = a/3 • 377.22 • 34.44 • 0.948 = 21.34/cIV

1.4.3 Example

Two loads are connected as shown in figure 1.17. Load 1 draws from the system Pl\ — 500 A; TV at
0.8 pf lagging, while the total load is St = lOOOkVA at 0.95 pf lagging. What is the pf of load 2?

12 THREE PHASE CIRCUITS AND POWER

r System

g

h

Fig. 1.17 Two loads fed from the same source

Note first that for the total load we can add real and reactive power for each of the two loads:

Pt = Pli+P L 2

Qt = QL1 + QL2

St t^ Sli + Sl2
From the information we have for the total load

P T = StpIt = 950/cW

Qt = S T sii^cos -1 0.95) = 312.25/cV At

Note positive Qt since pf is lagging

For the load LI, P L1 = 500kW, pfi = 0.8 lag,

Sli

500 • 10 3
0.8

625WA

L1 = JS 2 L1 -Pl 1 =37bkVAr

Ql\ is again positive, since pf is lagging.
Hence,

P L2 =P T - P L1 = 450/cW
)i2 = Qt~ Qli = -G2.75kVAr

(1-26)

and

pl

L2

450

L2

S L 2 V420 2 + 62.75 2

CALCULATIONS IN THREE-PHASE SYSTEMS 13

Notes

• A sinusoidal signal can be described uniquely by:

1. as e.g. v(t) = 5 sin(27r/£ + (j) v ),

2. by its graph,

3. as a phasor and the associated frequency.

one of these descriptions is enough to produce the other two. As an exercise, convert between
phasor, trigonometric expression and time plots of a sinusoid waveform.

• It is the phase difference that is important in power calculations, not phase. The phase alone of
a sinusoidal quantity does not really matter. We need it to solve circuit problems, after we take
one quantity (a voltage or a current) as reference, i.e. we assign to it an arbitrary value, often
0. There is no point in giving the phase of currents and voltages as answers, and, especially
for line-line voltages or currents in A circuits, these numbers are often wrong and anyway
meaningless.

• In both 3 -phase and 1 -phase systems the sum of the real power and the sum of the reactive
power of individual loads are equal respectively to the real and reactive power of the total load.
This is not the case for apparent power and of course not for power factor.

• Of the four quantities, real power, reactive power, apparent power and power factor, any two
describe a load adequately. The other two can be calculated from them.

• To calculate real reactive and apparent Power when using formulae 1.7, 1.10 1.11 we have to
use absolute not complex values of the currents and voltages. To calculate complex power
using 1.12 we do use complex currents and voltages and find directly both real and reactive
power.

• When solving a circuit to calculate currents and voltages, use complex impedances, currents
and voltages.

• Notice two different and equally correct formulae for 3 -phase power.

2

Magnetics

Chapter Objectives

In this chapter you will learn the following:

• How Maxwell's equations can be simplified to solve simple practical magnetic problems

• The concepts of saturation and hysteresis of magnetic materials

• The characteristics of permanent magnets and how they can be used to solve simple problems

• How Faraday's law can be used in simple windings and magnetic circuits

• Power loss mechanisms in magnetic materials

• How force and torque is developed in magnetic fields

2.1 INTRODUCTION

Since a good part of electromechanical energy conversion uses magnetic fields it is important early
on to learn (or review) how to solve for the magnetic field quantities in simple geometries and under
certain assumptions. One such assumption is that the frequency of all the variables is low enough
to neglect all displacement currents. Another is that the media (usually air, aluminum, copper, steel
etc.) are homogeneous and isotropic. We'll list a few more assumptions as we move along.

2.2 THE GOVERNING EQUATIONS

We start with Maxwell's equations, describing the characteristics of the magnetic field at low fre-
quencies. First we use:

V • B = (2.1)

15

16

MAGNETICS

the integral form of which is:

/

B • dA =

for any path. This means that there is no source of flux, and that all of its lines are closed.
Secondly we use

/H.a-/

3-dA

(2.2)

(2.3)

where the closed loop is defining the boundary of the surface A. Finally, we use the relationship
between H, the strength of the magnetic field, and B, the induction or flux density.

B = /z r /uoH

(2.4)

where /io is the permeability of free space, 47rlO~ 7 Tm/A, and fi r is the relative permeability of the
material, 1 for air or vacuum, and a few hundred thousand for magnetic steel.

There is a variety of ways to solve a magnetic circuit problem. The equations given above, along
with the conditions on the boundary of our geometry define a boundary value problem. Analytical
methods are available for relatively simple geometries, and numerical methods, like Finite Elements
Analysis, for more complex geometries.

Here we'll limit ourselves to very simple geometries. We'll use the equations above, but we'll add
boundary conditions and some more simplifications. These stem from the assumption of existence
of an average flux path defined within the geometry. Let's tackle a problem to illustrate it all. In

Fig. 2.1 A simple magnetic circuit

figure 2.1 we see an iron ring with cross section A c , average diameter r, that has a gap of length g
and a coil around it of N turns, carrying a current i. The additional assumptions we'll make in order
to calculate the magnetic field density everywhere are:

• The magnetic flux remains within the iron and a tube of air, the airgap, defined by the cross
section of the iron and the length of the gap. This tube is shown in dashed lines in the figure.

• The flux flows parallel to a line, the average flux path, shown in dash-dot.

THE GOVERNING EQUATIONS 1 7

• Flux density is uniform at any cross-section and perpendicular to it.
Following one flux line, the one coinciding with the average path, we write:

H -d\

J -dA

(2.5)

where the second integral extends over any surface (a bubble) terminating on the path of integration.
But equation 2.2, together with the first assumption assures us that for any cross section of the
geometry the flux, \$ = J* B dA = B avg A c , is constant. Since both the cross section and the flux
are the same in the iron and the air gap, then

and finally

l^iron-H-iron f^air-H-a

H-irony^ 1 ^

-g) + H

gap 9 ^ 1

(27rr -

_l~l j iron

-g) + g

H — Ni

11 gap — 1 v l

(2.6)

H

yi

H

y2

i

1

i

\ V

e V

(

)

(

n
C

I

u

'

. y

. y

Fig. 2.2 A slightly complex magnetic circuit

Let us address one more problem: calculate the magnetic field in the airgap of figure 2.2,
representing an iron core of depth d . Here we have to use two loops like the one above, and we have

18 MAGNETICS

a choice of possible three. Taking the one that includes the legs of the left and in the center, and the
outer one, we can write:

Ht-l + Hyi-y + Hc

c + H g - g + H c - c + H y i • y
Hi-h + 2H yl -y + H r -l + 2H y2 • y

Ni
Ni

(2.7)

Applying equation 2.3 to the closed surface shown shaded we also obtain:

B\ A y — B c A c — B r A y =

and of course

Bi = /i Hi, B c = [i H C1 B r = [i H r B g = ii\$H g

The student can complete the problem. We notice though something interesting: a similarity
between Kirchoff 's equations and the equations above. If we decide to use:

n

T

B A

I
Aii

Ni

(2.8)

(2.9)

(2.10)

then we notice that we can replace the circuits above with the one in figure 2.3, with the following
correspondence:

Ryl

i?i

W^

Ry 2

k*'

w^

Fig. 2.3 Equivalent electric circuit for the magnetic circuit in figure 2.2

Magnetic

Electrical

J 7 , magnetomotive force

<E>, flux

11, reluctance

V, voltage, or electromotive force

/, current

R, resistance

SATURATION AND HYSTERESIS

19

This is of course a great simplification for students who have spent a lot of effort on electrical
circuits, but there are some differences. One is the nonlinearity of the media in which the magnetic
field lives, particularly ferrous materials. This nonlinearity makes the solution of direct problems a
little more complex (problems of the type: for given flux find the necessary current) and the inverse
problems more complex and sometimes impossible to solve without iterations (problems of the type:
for given currents find the flux).

2.3 SATURATION AND HYSTERESIS

Although for free space a equation 2.3 is linear, in most ferrous materials this relationship is nonlinear.
Neglecting for the moment hysteresis, the relationship between H and B can be described by a curve
of the form shown in figure 2.4. From this curve, for a given value of B or H we can find the other
one and calculate the permeability /j, = B/H.

Fig. 2.4 Saturation in ferrous materials

In addition to the phenomenon of saturation we have also to study the phenomenon of hysteresis
in ferrous materials. The defining difference is that if saturation existed alone, the flux would be a
unique function of the field intensity. When hysteresis is present, flux density for a give value of
field intensity, H depends also on the history of magnetic flux density, B in it. We can describe the
relationship between field intensity, H and flux density B in homogeneous, isotropic steel with the
curves of 2.5. These curves show that the flux density depends on the history of the magnetization of
the material. This dependence on history is called hysteresis. If we replace the curve with that of the
locus of the extrema, we obtain the saturation curve of the iron, which in itself can be quite useful.

Going back to one of the curves in 2.5, we see that when the current changes sinusoidally between
the two values, i and — i, then the point corresponding to (H,B) travels around the curve. During
this time, power is transferred to the iron, referred to as hysteresis losses, Physt- The energy of these
losses for one cycle is proportional to the area inside the curve. Hence the power of the losses is
proportional to this surface, the frequency, and the volume of iron; it increases with the maximum
value of B\

Physt = kfB x Kx<2 (2.11)

20

MAGNETICS

Fig. 2.5 Hysteresis loops and saturation

If the value of H, when increasing towards H, does so not monotonously, but at one point, Hi,
decreases to H 2 and then increases again to its maximum value, H, a. minor hysteresis loop is created,
as shown in figure 2.6. The energy lost in one cycle includes these additional minor loop surfaces.

Fig. 2.6 Minor loops on a hysteresis curve

PERMANENT MAGNETS 21

Fig. 2.7 Hysteresis curve in magnetic steel

2.4 PERMANENT MAGNETS

If we take a ring of iron with uniform cross section and a magnetic characteristic of the material
that in figure 2.7, and one winding around it, and look only at the second quadrant of the curve, we
notice that for H = 0, i.e. no current in an winding there will be some nonzero flux density, B r . In
addition, it will take current in the winding pushing flux in the opposite direction (negative current)
in order to make the flux zero. The iron in the ring has became a permanent magnet. The value
of the field intensity at this point is — H c . In practice a permanent magnet is operating not at the
second quadrant of the hysteresis loop, but rather on a minor loop, as shown on figure 2.6 that can
be approximated with a straight line. Figure 2.8 shows the characteristics of a variety of permanent
magnets. The curve of a permanent magnet can be described by a straight line in the region of
interest, 2.9, corresponding to the equation:

Bm =

Hr,

H r

H r

- B r

(2.12)

2.4.1 Example

In the magnetic circuit of figure 2.10 the length of the magnet is l m = lcm, the length of the air gap
is g = lmm and the length of the iron is U = 20cm. For the magnet B r = LIT, H c = 750k A/ m.
What is the flux density in the air gap if the iron has infinite permeability and the cross section is
uniform?

22

MAGNETICS

n:

Nku>

&yn\\

Uku

/

As

Aim"

CO

kCol

salt

7

CJ

Cert

mic

f

//

/

/

®

/

®

7

(A)

/

/

/

/

/

/

/

/

/

y

/

f

1.4

B

(T)

0.6

0.4

12 10 8 G 4 2 0"

-H (kA/m)

Fig. 2.8 Minor loops on a hysteresis curve

Since the cross section is uniform, B is the same everywhere, and there is no current:

Hi • 0.2 + H g -g + H m • k =

for infinite iron permeability Hi = 0, hence,

B a%r —g + (B m - 1.1) ( ^ J /, =

=> B • 795.77 + (B - 1.1) • 6818 =
B = 0.985T

We'll see now how voltage is generated in a coil and the effects it may have on a magnetic material.
This theory, along with the previous chapter, is essential in calculating the transfer of energy through
a magnetic field.

First let's start with the governing equation again. When flux through a coil changes for whatever
reason (e.g. change of the field or relative movement), a voltage is induced in this coil. Figure 2.1 1

A B

7

/ R

-/

m

H

m

H

Fig. 2.9 Finding the flux density in a permanent magnet

\

/

/

/

m

\

t

Fig. 2. 10 Magnetic circuit for Example 2.4. 1

shows such a typical case. Faraday's law relates the electric and magnetic fields. In its integral form:

d

fe-di = -±[

JC dt J A

S-dl=-— / B dA

IC dt J A

(2.13)

24 MAGNETICS

and in the cases we study it becomes:

v(t)

d\$(t)
dt

(2.14)

Fig. 2.11 Flux through a coil

If a coil has more than one turns in series, we define as flux linkages of the coil, A, the sum of the
flux through each turn,

* = £*<

(2.15)

and then:

v(t)

d\(t)

dt

(2.16)

2.5.1 Example

For the magnetic circuit shown below \±i ron = \i Q • 10 5 , the air gap is lmm and the length of the iron
core at the average path is lm. The cross section of the iron core is 0.04m 2 . The winding labelled
'primary ' has 500 turns. A sinusoidal voltage of 60Hz is applied to it. What should be the rms
value of it if the flux density in the iron (rms) is O.ST? What is the current in the coil? The voltage
induced in the coil will be

But if B(t) = B sin(27r/t) => \$(t) = AB sm(2irft)
=> \$(t) = 0.04(^2 • 0.8) sin(377£) Wb

ei(t)

ei(t) = 500 0.04^2 • 0.8 • 377sin(377£ + |)

dt

V

Ei

gi

V2

500 • 0.04 • 0.8 • 377 = 6032V

EDDY CURRENTS AND EDDY CURRENT LOSSES 25

c_

c

D

~)

C

c

D

^

~^

v

_J

prirr

lary

secondary

F/gf. 2. 72 Magnetic circuit for Example 2.5. 1
7b calculate the current we integrate around the loop of the average path:

biron = Bair = V2 • 0.8 sin(377£) =4> H air = s\n(?>77t)A/m

Hj A

72 -0.8

sin(377£)A/ra

Finally

500 • z

72-0.8 8111(377^) f 1 1-HT 3
10^ + 1

z = 1.819 sin(377t)A =>/= — = 1.2S6A

v2

2.6 EDDY CURRENTS AND EDDY CURRENT LOSSES

When the flux through a solid ferrous material varies with time, currents are induced in it. Figure
2.13 gives a simple explanation: Let's consider a ring of iron defined within the material shown in
black and a flux <1> through it, shown in grey. As the flux changes, a voltage e = d<&/dt is induced
in the ring. Since the ring is shorted, and has a resistance R, a current flows in it, and Joule losses,
Peddy = e 2 / R, result. We can consider a multitude of such rings in the material, resulting into Joule
losses, but the method discussed above is not the appropriate one to calculate these losses. We can,
though, estimate that for sinusoidal flux, the flux, voltage, and losses are:

26 MAGNETICS

Fig. 2.13 Eddy currents in solid iron

<E> = <f> sm(ou t) = AB sm(cu t)
e = uj& cos(uj t) = 2ttA/B cos(uj t)

eddy = % J B

(2.17)
(2.18)
(2.19)

which tells us that the losses are proportional to the square of both the flux density and frequency.
A typical way to decrease losses is to laminate the material, as shown in figure 2.14, decreasing the
paths of the currents and the total flux through them.

Iron

insulation

Fig. 2. 14 Laminated steel

TORQUE AND FORCE 27

2.7 TORQUE AND FORCE

Calculating these is quite more complex, since Maxwell's equations do not refer directly to them.
The most reasonable approach is to start from energy balance. Then the energy in the firles Wf is
the sum of the energy that entered through electrical and mechanical sources.

This in turn can lead to the calculation of the forces since

K J

Y, fkdx k = J2 e 3h dt ~ dW f ( 2 ' 21 )

fc=l 3 = 1

Hence for a small movement, dxk, the energies in the equation should be evaluated and from
these, forces (or torques), /&, calculated.

Alternatively, although starting from the same principles, one can use the Maxwell stress tensor
to find forces or torques on enclosed volumes, calculate forces using the Lorenz force equation, here
F = UB, or use directly the balance of energy. Here we'll use only this last method, e.g. balance
the mechanical and electrical energies.

In a mechanical system with a force F acting on a body and moving it at velocity v in its direction,
the power Pmech is

Pmech =F-V (2.22)

This eq. 2.22, becomes for a rotating system with torque T, rotating a body with angular velocity

^mech-

-Lmech -*- ' ^mech \Z.Zj)

On the other hand, an electrical source e, supplying current i to a load provides electrical power

-*eZec

Peiec = e-i (2.24)

Since power has to balance, if there is no change in the field energy,

■Lelec = -Lmech ^ J- ' ^mech = & ' 1 yZ.Zj)

Notes

• It is more reasonable to solve magnetic circuits starting from the integral form of Maxwell's
equations than finding equivalent resistance, voltage and current. This also makes it easier to
use saturation curves and permanent magnets.

• Permanent magnets do not have flux density equal to Br. Equation 2.12defines the relation
between the variables, flux density B m and field intensity H m in a permanent magnet.

• There are two types of iron losses: eddy current losses that are proportional to the square of
the frequency and the square of the flux density, and hysteresis losses that are proportional to
the frequency and to some power x of the flux density.

3

Transformers

Although transformers have no moving parts, they are essential to electromechanical energy conver-
sion. They make it possible to increase or decrease the voltage so that power can be transmitted at
a voltage level that results in low costs, and can be distributed and used safely. In addition, they can
provide matching of impedances, and regulate the flow of power (real or reactive) in a network.

In this chapter we'll start from basic concepts and build the equations and circuits corresponding
first to an ideal transformer and then to typical transformers in use. We'll introduce and work with
the per unit system and will cover three-phase transformers as well.

After working on this chapter, you'll be able to:

• Choose the correct rating and characteristics of a transformer for a specific application,

• Calculate the losses, efficiency, and voltage regulation of a transformer under specific operating
conditions,

Experimentally determine the transformer parameters given its ratings.

3.1 DESCRIPTION

When we see a transformer on a utility pole all we see is a cylinder with a few wires sticking out.
These wires enter the transformer through bushings that provide isolation between the wires and
the tank. Inside the tank there is an iron core linking coils, most probably made with copper, and
insulated. The system of insulation is also associated with that of cooling the core/coil assembly.
Often the insulation is paper, and the whole assembly may be immersed in insulating oil, used to
both increase the dielectric strength of the paper and to transfer heat from the core-coil assembly to
the outer walls of the tank to the air. Figure 3.1 shows the cutout of a typical distribution transformer

29

30 TRANSFORMERS

HV bushing

Surge suppressor

Coil

Fig. 3. 1 Cutaway view of a single phase distribution transformer. Notice only one HV bushing and lightning
arrester

3.2 THE IDEAL TRANSFORMER

Few ideal versions of human constructions exist, and the transformer offers no exception. An ideal
transformer is based on very simple concepts, and a large number of assumptions. This is the
transformer one learns about in high school.

Let us take an iron core with infinite permeability and two coils wound around it (with zero
resistance), one with N\ and the other with 7V 2 turns, as shown in figure 3.2. All the magnetic flux is
to remain in the iron. We assign dots at one terminal of each coil in the following fashion: if the flux

~<Pr7

Fig. 3.2 Magnetic Circuit of an ideal transformer

in the core changes, inducing a voltage in the coils, and the dotted terminal of one coil is positive
with respect its other terminal, so is the dotted terminal of the other coil. Or, the corollary to this,
current into dotted terminals produces flux in the same direction.

THE IDEAL TRANSFORMER 31

Assume that somehow a time varying flux, \$(£), is established in the iron. Then the flux linkages
in each coil will be Ai = Ni\$(t) and A2 = N 2 &(t). Voltages will be induced in these two coils:

ei(t)

e 2 (t)

and dividing:

~dt

■ =

m

d<S>
~dt

d\ 2
dt

■ =

N 2

d<£>
~dt

ei(t)

m

e 2 (t)

N 2

(3.1)

(3.2)

(3.3)

On the other hand, currents flowing in the coils are related to the field intensity H. If currents
flowing in the direction shown, i\ into the dotted terminal of coil 1, and i 2 out of the dotted terminal
of coil 2, then:

N 1 -i 1 (t)-N 2 i 2 (t)=H-l (3.4)

but B = j^ironH, and since B is finite and /ii ron is infinite, then H = 0. We recognize that this is
practically impossible, but so is the existence of an ideal transformer.
Finally:

i 2 TVi

Equations 3.3 and 3.5 describe this ideal transformer, a two port network. The symbol of a
network that is defined by these two equations is in the figure 3.3. An ideal transformer has an

N 1

N,

Fig. 3.3 Symbol for an ideal transformer

interesting characteristic. A two-port network that contains it and impedances can be replaced by an
equivalent other, as discussed below. Consider the circuit in figure 3.4a. Seen as a two port network

(a)

(b)

Fig. 3.4 Transferring an impedance from one side to the other of an ideal transformer

32 TRANSFORMERS

with variables v±, i\, V2, i<i, we can write:

ei

= U\ — i\Z

e 2

N 2 N 2 N 2 . „
= — ei = — ua i\Z

^2

Af 2 ^2 . /JV 2

(3.6)

(3.7)

(3.8)

which could describe the circuit in figure 3.4b. Generally a circuit on a side 1 can be transferred to
side 2 by multiplying its component impedances by (N2/N1) 2 , the voltage sources by (N2/N1) and
the current sources by (N1/N2), while keeping the topology the same.

3.3 EQUIVALENT CIRCUIT

To develop the equivalent circuit for a transformer we'll gradually relax the assumptions that we had
first imposed. First we'll relax the assumption that the permeability of the iron is infinite. In that
case equation 3.4 does not revert to 3.5, but rather it becomes:

Nxh - N 2 i 2 = n\$ ri

(3.9)

where 1Z is the reluctance of the path around the core of the transformer and <£ m the flux on this path.
To preserve the ideal transformer equations as part of our new transformer, we can split i\ to two
components: one i[, will satisfy the ideal transformer equation, and the other, i\^ ex will just balance
the right hand side. Figure 3.5 shows this.

U

+ ^

ideal transformer
Fig. 3.5 First step to include magnetizing current

e2

H

*1 + 1l,ex

H-l

(3.10)
(3.11)
(3.12)

EQUIVALENT CIRCUIT 33

We can replace the current source, i\, ex , with something simpler if we remember that the rate of
change of flux <£ m is related to the induced voltage e\ :

ei

d\$ r ,

dt
d{N 1 i^ ex /ll)

dt

n dt

(3.13)
(3.14)
(3.15)

Since the current i\^ ex flows through something, where the voltage across it is proportional to its
derivative, we can consider that this something could be an inductance. This idea gives rise to the

N 2

equivalent circuit in figure 3.6, where L m = -£ Let us now relax the assumption that all the flux has

n

12

1, ex

ideal transformer
Fig. 3.6 Ideal transformer plus magnetizing branch

to remain in the iron as shown in figure 3.7. Let us call the flux in the iron <E> m , magnetizing flux, the
flux that leaks out of the core and links only coil 1, &n, leakage flux 1, and for coil 2, <E>/ 2 , leakage
flux 2. Since <E>q links only coil 1, then it should be related only to the current there, and the same
should be true for the second leakage flux.

Primary windings

Secondary
windings

Flux lines caused
by the winding

currents

Fig. 3.7 If the currents in the two windings were to have cancelling values of N • i, then the only flux left
would be the leakage fluxes. This is the case shown here, designed to point out these fluxes.

34

TRANSFORMERS

^ ll =N 1 i 1 /7Z ll
\$12 = N 2 i 2 /lli 2

(3.16)
(3.17)

where IZn and 1Z\ 2 correspond to paths that are partially in the iron and partially in the air. As these
currents change, so do the leakage fluxes, and a voltage is induced in each coil:

ei

e 2

dAi =N ( d& ni
dt 1 \dt

•TVi

dt

ei

d\ 2
~dt

N 2

d<S>rr,

dt

AT d\$ '2

N 2 —j— = e 2
dt

Nl \ dh
Wi) ~dt
Nl\ dh

n i2 ) dt

(3.18)
(3.19)

If we define La

-,£,

72

— , then we can arrive to the equivalent circuit in figure 3.8. To this

-12

12

°2

hi

Ni N2

Ideal transformer
Fig. 3.8 Equivalent circuit of a transformer plus magnetizing and leakage inductances

v 2

1. The winding (ohmic) resistance in each coil, Ri, w d g , R2,wdg, with losses Pi, w d g = iiRi,wdg,

°22,wdg

Pll.inda — i 2 R2,wdg> an ^

2. some resistance to represent iron losses. These losses (at least the eddy-current ones) are
proportional to the square of the flux. But the flux is proportional to the square of the induced
voltage ei, hence Pi ron = ke\. Since this resembles the losses of a resistance supplied by
voltage ei, we can develop the equivalent circuit 3.9.

3.3.1 Example

Let us now use this equivalent circuit to solve a problem. Assume that the transformer has a
turns ratio of 4000/120, with R\, w d g = 1.60, R 2 , w dg = 1.44raO, Lq = 21mH, L\ 2 = 19/xi^,
R c = 160/cO, L m = 450H. assume that the voltage at the low voltage side is 60Hz, V 2 = 1201^,
and the power there is P 2 = 20kW, at pf = 0.85 lagging. Calculate the voltage at the high voltage
side and the efficiency of the transformer.

X m = L m * 2tt60 = 169.7/cO

X x = 7.920

X 2 = 7.16mO

EQUIVALENT CIRCUIT 35

R wdg,1 ^
v 1 L I2 R

'1 i\

) H.«4l +

Ideal transformer

-ww-

R wdg, 2

v.

Ni:N 2

(a)

Fig. 3.9 Equivalent circuit for a real transformer

xZ-31.8°

196.1336 Z " 31 - 8 °A

I 2 = P L /(V L pf)
E 2 = V 2 + I 2 (^^, 2 + jX, 2 ) = 120.98 + J1M5V

Ei

X 2

11 " Ui

II

Ei

1

1

E 2 = 4032.7 + J34.83V

I 2 = 5.001 -J3.1017A

0.0254 -J0.0236A

Re jx n

Ii = Ii,ex + Ii = 5.0255 - J3.125A
Vi = Ei + Ii (^^,i + jX u ) = 4065.5 + J69.2F = 4066 Zo - 9 V

77z£ power losses are concentrated in the windings and core:

P wd9 ,2 = llR wd9 ,2 = 196.13 2 • 1.44 • 10"

55.39W

Pwd 9 ,i = IiRwdg,i = 5.918 2 • 1.6 = 56.04W

Pcore = El/ R c = 4032.8 2 /(i60 . io 3 ) = 101.64W

Ploss — Pwdg,l + Pwdg,2 + Pcore = 213. OSW
Pout Pout WkW

(Pout + Ploss) 20kW + 213.08W

0.9895

36 TRANSFORMERS

3.4 LOSSES AND RATINGS

Again for a given frequency, the power losses in the core (iron losses) increase with the voltage
ei (or e 2 ). These losses cannot be allowed to exceed a limit, beyond which the temperature of the
hottest spot in the transformer will rise above the point that will decrease dramatically the life of the
insulation. Limits therefore are put to E\ and E 2 (with a ratio of Ni/N 2 ), and these limits are the
voltage limits of the transformer.

Similarly, winding Joule losses have to be limited, resulting in limits to the currents I\ and I 2 .

Typically a transformer is described by its rated voltages, Ein and E 2 at, that give both the limits
and turns ratio. The ratio of the rated currents, Iin/Iin,'^ the inverse of the ratio of the voltages
if we neglect the magnetizing current. Instead of the transformer rated currents, a transformer is
described by its rated apparent power:

Sn = E\nI\n = E 2 nI 2 n (3.20)

Under rated conditions, i.e. maximum current and voltage, in typical transformers the magnetizing
current I\^ ex , does not exceed 1% of the current in the transformer. Its effect therefore on the voltage
drop on the leakage inductance and winding resistance is negligible.

Under maximum (rated) current, total voltage drops on the winding resistances and leakage
inductances do not exceed in typical transformers 6% of the rated voltage. The effect therefore of
the winding current on the voltages E\ and E 2 is small, and their effect on the magnetizing current
can be neglected.

These considerations allow us to modify the equivalent circuit in figure 3.9, to obtain the slightly
inaccurate but much more useful equivalent circuits in figures 3.10a, b, and c.

3.4.1 Example

Let us now use these new equivalent circuits to solve the previous problem 3.3.1. We } ll use the circuit
in 3.10b. Firs let's calculate the combined impedances:

Rwdg = Rwdg,l + ( "T^- ) Rwdg,2 = 3.21]

X t = X lA + f^\ X h2 = 15.87591]

then, we solve the circuit:

la = PL/(V L -pf) l - 31 - 8 ° = 196.1336 Z - 31 - 8 °A

E2 = V2

Ii=I 2 -(^|) =5+j3.102A

N 2

4000V

Ii.es = Ei ( -J- + -^r-) = 0.0258 - J0.0235A
\R C jX m )

Ii = Ii,ex + Ii = 5.0259 - J3.125A
Vi = Ei + I'i (Rwdg + jXi) = 4065 + J69.45V = 4065 Zl V

PER-UNIT SYSTEM 37

: 1

l 2

Ni=N 2
R wdg,l L ll R wdg,2 L 12

Vi c

hn

+ +

e 2 v 2

V l

+

R wdg,l L ll R wdg,2 L 12

Ni=N 2

^

: 2

+

Ni:N 2

Fig. 3.10 Simplified equivalent circuits of a transformer

The power losses are concentrated in the windings and core:

P wdg = I[R wdg = 110.79W

Pcore = V?/R c = 103.32W

Ploss = Pwdg + Pcore = 214.11W

-^out

-LOUt

20fcW nnnni

Pin (Pout + Ploss) 20kW + 221.411W

3.5 PER-UNIT SYSTEM

The idea behind the per unit system is quite simple. We define a base system of quantities, express
everything as a percentage (actually per unit) of these quantities, and use all the power and circuit
equations with these per unit quantities. In the process the ideal transformer in 3.10 disappears.

Working in p.u. has a some other advantages, e.g. the range of values of parameters is almost the
same for small and big transformers.

Working in the per unit system adds steps to the solution process, so one hopes that it simplifies
the solution more than it complicates it. At first attempt, the per unit system makes no sense. Let us
look at an example:

38 TRANSFORMERS

3.5.1 Example

A load has impedance 10 + jhQ, and is fed by a voltage of 100V. Calculate the current and power

Solution 1 the current will be

I L = ^ = -^ = 8.94^-°A
Z L 10 + j5

and the power will be

Pl = V L I L • pf = 100 • 8.94 • cos(26.57) = 800^
Solution 2 Let's use the per unit system.

1. define a consistent system of values for base. Let us choose Vb = hOV, lb = 10A This means
that Z h = V b /h = 5fi, and P b = V b • I b = 500W, Q b = 500V Ar, S b = 500V A.

2. Convert everything to pu. Vl, pu — Vl/B^ = 2pu, Zl^ pu = (10 + j5)/5 = 2 + jl pu.

3. solve in the pu system.

Il.pu = ^ = x^- = 0.894 Z - 26 - 57 ° pu

Pl, P u = V LpU I L , pu -pf = 2 • 0.894 • cos(26.57°) = 1.6 pu

4. Convert back to the SI system

II = h, pu • h = 0.894 • 10 = 8.94A
Pi = Pl, pu • Pb = 1.6 • 500 = 800W

The second solution was a bit longer and appears to not be worth the effort. Let us now apply this
method to a transformer, but be shrewder in choosing our bases. Here we'll need a base system for
each side of the ideal transformer, but in order for them to be consistent, the ratio of the voltage and
current bases should satisfy:

\L , AT.

(3.21)

(3.22)
=* S lb = V lb I lb = V 2h hb = S 2b (3.23)

i.e. the two base apparent powers are the same, as are the two base real and reactive powers.

We often choose as bases the rated quantities of the transformer on each side, This is convenient,
since the transformer most of the time operates at rated voltage (making the pu voltage unity), and
the currents and power are seldom above rated, above lpu.

Notice that the base impedances on the two sides are related:

Zi, b = ^ (3.24)

Z% „ . £>_(«0> (3.25,

-'2.6 = Ittj ^l.b (3.26)

v lb

Ni

v 2b

N 2

lib

N 2

hb

~ JVi

Vl,b

h,b

v 2 , b

(N 2 ^

\ 2 Vi, b

hb ~

"Uiy

1 h,b

2

) Zijb

PER-UNIT SYSTEM 39

We notice that as we move impedances from the one side of the transformer to the other, they get
multiplied or divided by the square of the turns ratio, ( ^ ) , but so does the base impedance, hence
the pu value of an impedance stays the same, regardless on which side it is.

Also we notice, that since the ratio of the voltages of the ideal transformer is Ei/E 2 = N1/N2,
is equal to the ratio of the current bases on the two sides on the ideal transformer, then

Ei ?P u = E 2 , pu
and similarly,

Il,pu — ^2,pu

This observation leads to an ideal transformer where the voltages and currents on one side are
identical to the voltages and currents on the other side, i.e. the elimination of the ideal transformer,
and the equivalent circuits of fig. 3.11 a, b. Let us solve again the same problem as before, with

R wdg,l L ll R wdg,2 L 12 ^ -L +

R c ha

(a) (b)

Fig. 3.11 Equivalent circuits of a transformer in pu

3.5.2 Example

A transformer is rated 30WA, 4000V/120V, with R w d g ,i = 1.6fl, R w d g ,2 = 1.44rafi, L tl =
21mH, L12 = 19/liH, R c = 160kQ, L m = 450H. The voltage at the low voltage side is 60Hz,
V2 = 120 V, and the power there is P2 = 20kW, atpf = 0.85 lagging. Calculate the voltage at the
high voltage side and the efficiency of the transformer.

1. First calculate the impedances of the equivalent circuit:

V lb = 4000V
S lb = 30kVA

30 • 1Q3 wa

hb = T^ = 7 ' bA

V 2
Z lb = -^ = 533ft

>->16

V 2b = 120V

,526 = S lb = 30kVA

I 2b = ^=250A

V2b

v 2b

Z 2b = -^ = 0.48fi

hb

40 TRANSFORMERS

2. Convert everything to per unit: First the parameters:

Rwdg,l,pu = Rwdg,l/Z\b = 0.003 pu
Rwdg,2,pu = Rwdg, 2I %2b = 0.003 pu

Xii lPU = — ~ = 0.015 pu

£\b

X l2 , pu = 2 -^= 0.0149 pu

^26

Rc lP u = -tt~ = 30 ° P u

^ lb

27r60L Zm

Xm,pu = ~ = 318p^

^Ib

V 2
V 2 , P u = — = lpu

V2b
P2

P 2 , pu = -&- = 0.6667pu

^2b

3. Solve in the pu system. We 7/ drop the pu symbol from the parameters and variables:

/ p \ /-arccos(pf)

h = (yr-ri) = 0.666 -jo.mpu

Vi = V 2 + I [R wdgA + R wdga + j(X n + X l2 )] = 1.0172 + jO.OlSSpu
lm = ^ + ^~ = 0.0034 - jO.OOSlpu

Re jX m

Ii = I m + I 2 = 0.06701- jOA16pu

Pwdg = l\ (RwdgA + Rewg,2) = 0.0037 fm
Pcore = -£-= 0.0034^

7] = ^ — = 0.9894

-Lwdg ~r -Lcore ~r i 2

4. Convert back to SI. The efficiency, r], is dimensionless, hence it stays the same. The voltage,
Vi is

V 1 =V 1 , pn F 16 =4069 Zl V

3.6 TRANSFORMER TESTS

We are usually given a transformer, with its frequency, power and voltage ratings, but without the
values of its impedances. It is often important to know these impedances, in order to calculate voltage
regulation, efficiency etc., in order to evaluate the transformer (e.g. if we have to choose from many)
or to design a system. Here we'll work on finding the equivalent circuit of a transformer, through
two tests. We'll use the results of these test in the per-unit system.

First we notice that if the relative values are as described in section 3.4, we cannot separate the
values of the primary and secondary resistances and reactances. We will lump R\, w dg and R2,wdg

TRANSFORMER TESTS 41
together, as well as X\\ and Xi 2 . This will leave four quantities to be determined, R w d g , X\, R c and

Xm-

3.6.1 Open Circuit Test

We leave one side of the transformer open circuited, while to the other we apply rated voltage (i.e.
V oc = lpu) and measure current and power. On the open circuited side of the transformer rated
voltage appears, but we just have to be careful not to close the circuit ourselves. The current that
flows is primarily determined by the impedances X m and R c , and it is much lower than rated. It is
reasonable to apply this voltage to the low voltage side, since (with the ratings of the transformer in
our example) is it easier to apply 120V\ rather than 4000 V\ We will use these two measurements to
calculate the values of R c and X m .

Dropping the subscript pu, using the equivalent circuit of figure 3.1 lb and neglecting the voltage
drop on the horizontal part of the circuit, we calculate:

(3.27)

p —

1 oc —
-1-oc

V 2 1

v oc - 1 -

R c R c

Voc ! Voc

Rc jX m

J-oc =

J ' + *

V R c x m

(3.28)

Equations 3.27 and 3.28, allow us to use the results of the short circuit test to calculate the vertical
(core) branch of the transformer equivalent circuit.

3.6.2 Short Circuit Test

To calculate the remaining part of the equivalent circuit, i.e the values of R w d g and Xi, we short
circuit one side of the transformer and apply rated current to the other. We measure the voltage of
that side and the power drawn. On the other side, (the short-circuited one) the voltage is of course
zero, but the current is rated. We often apply voltage to the high voltage side, since a) the applied
voltage need not be high and b) the rated current on this side is low.
Using the equivalent circuit of figure 3.1 la, we notice that:

-*sc -L sc-^wdg J- ' -t^wdg \P'^)

V sc = I sc (R w dg + jXi)

Vsc = 1-y/Rl^ + Xf (3.30)

Equations 3.29 and 3.30 can give us the values of the parameters in the horizontal part of the
equivalent circuit of a transformer.

3.6.1 Example

A 60Hz transformer is rated 30kVA, 4000F/120V\ The open circuit test, performed with the high
voltage side open, gives P oc = 100W, I oc = 1.1455 A The short circuit test, performed with the
low voltage side shorted, gives P sc = 180 W, V sc = 129. 79F . Calculate the equivalent circuit of
the transformer in per unit.

42 TRANSFORMERS

First define bases:

hb

Z\b

hb

%2b

4000V

30kVA

30 • 10 3

4-10 3

7.5A

V 2

-^ = 533(7

&lb

120V

S lb = 30kVA

Sib

v 2b

Vi±

hb

250A
0.4817

Convert now everything to per unit:

V M

180

30 • 10 3
129.79

4000
100

30 • 10 3
1.1455

250

= 0.006ppu
= 0.0324prx
= 0.003333p^
= 0.0046p^

Let's calculate now, dropping the pu subscript:

p

1 sc
|Vsc|

T z 7?

1 sc- rL wd

-^wdg ■'■sc/^sc ' S(

\I sc \-\R wdg +jX l \ = l.^Rl t

V 2 l 2

-f => R c = — = 300pu

*oc * oc /J- J-

~r7 JxZ ~y& c xl

■*?

0.006pw
=* X t = Jv? c - Rl

Xn

0.0318p^

he Rl

3\%pu

A more typical problem is of the type:

3.6.2 Example

A 60Hz transformer is rated 30k V A, 4000 V/120V. Its short circuit impedance is 0.032Apu and the
open circuit current is 0.0046pu. The rated iron losses are 100 W and the rated winding losses are
1801V. Calculate the efficiency and the necessary primary voltage when the load at the secondary
is at rated voltage, 20kW at O.Spf lagging.

THREE-PHASE TRANSFORMERS 43

Working in pu:

Z sc = 0.0324p^

1 80

P S c = Rwdg = 3Q 3 = 6 • 10~ 3 pU

X l = Jzl-Rl d =0M7pu

p ° c = h * Rc = h = loo/so ■ io3 = 300pM

p ° c ~h c = 318pu

Having finished with the transformer data, let us work with the load and circuit. The load power
is 20kW, hence:

20 • 10 3

P2 = SOTIb^ = ° MQ7PU
but the power at the load is:

P 2 = V 2 I 2 pf => 0.6667 = 1 • I 2 • 0.8 => I 2 = 0.8333p^
Then to solve the circuit, we work with phasors. We use the voltage V 2 as reference:
V 2 = V 2 = lpu

I 2 = 0.8333 Zcos_1 °- 8 = 0.6667 - j0.5pu

Vi = V 2 + I 2 (Rwdg + jXi) = 1.0199 + j0.00183p^ => V x = 1.02pu
Pwd 9 = Ii-R wdg =0.0062pu
P c = V?/R c = 0.034pu

71 = P +P 2 +P = °- 986
Finally, we convert the voltage to SI

Vi = V liPU • V bl = 1.021 • 4000 = 4080V

3.7 THREE-PHASE TRANSFORMERS

We'll study now three-phase transformers, considering as consisting of three identical one-phase
transformers. This method is accurate as far as equivalent circuits and two-port models are our
interest, but it does not give us insight into the magnetic circuit of the three-phase transformer. The
primaries and the secondaries of the one-phase transformers can be connected either in A or in Y.
In either case, the rated power of the three-phase transformer is three times that of the one-phase
transformers. For A connection,

Vu = V H (3.31)

// = VSht (3.32)

For Y connection

V lt = V3V 1(P (3.33)

Ii = h* (3.34)

44 TRANSFORMERS

h

V1

V1

"2

V 2

V 2

V 2

V1

Vi

V 2

v 2

v 2

(a)

(b)

Fig. 3.12 Y — Y and Y — A Connections of three-phase Transformers

V1

V1

V1

v 2

v 2

v 2

Vi

V1

V1

v 2

v 2

v 2

(a)

(b)

F/flf. 3. 13 A — Y and A — A Connections of three-phase Transformers

3.8 AUTOTRANSFORMERS

An autotransformer is a transformer where the two windings (of turns N\ and A^ 2 ) are not isolated
from each other, but rather connected as shown in figure 3.14. It is clear form this figure that the

voltage ratio in an autotransformer is

while the current ratio is

Vi

N x +N 2

v 2

N 2

h

N!+N 2

h

N 2

AUTOTRANSFORMERS 45

(3.35)

(3.36)

The interesting part is that the coil of turns of N\ carries current I\ , while the coil of turns N 2 carries
the (vectorial) sum of the two currents, Ii — 12. So if the voltage ratio where 1, no current would flow
through that coil. This characteristic leads to a significant reduction in size of an autotransformer
compared to a similarly rated transformer, especially if the primary and secondary voltages are of
the same order of magnitude. These savings come at a serious disadvantage, the loss of isolation
between the two sides.

"1

N 1

Vi

'2

+

i,-',n "> 2

Fig. 3. 14 An Autotransformer

Notes

• To understand the operation of transformers we have to use both the Biot-Savart Law and

• Most transformers operate under or near rated voltage. As the voltage drop in the leakage
inductance and winding resistances are small, the iron losses under such operation transformer
are close to rated.

• The open- and short-circuit test are just that, tests. They provide the parameters that define the
operation of the transformer.

• Three-phase transformers can be considered to be made of three single-phase transformers for
the purposes of these notes. The main issue then is to calculate the ratings and the voltages
and currents of each.

46 TRANSFORMERS

• Autotransformers are used mostly to vary the voltage a little. It is seldom that an autotrans-
former will have a voltage ratio greater than two.

4

Concepts of Electrical
Machines; DC motors

DC machines have faded from use due to their relatively high cost and increased maintenance
requirements. Nevertheless, they remain good examples for electromechanical systems used for
control. We'll study DC machines here, at a conceptual level, for two reasons:

1 . DC machines although complex in construction, can be useful in establishing the concepts of
emf and torque development, and are described by simple equations.

2. The magnetic fields in them, along with the voltage and torque equations can be used easily to
develop the ideas of field orientation.

In doing so we will develop basic steady- state equations, again starting from fundamentals of the
electromagnetic field. We are going to see the same equations in 'Brushless DC motors, when we
discuss synchronous AC machines.

4.1 GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS

This geometry describes an outer iron window (stator), through which (i.e. its center part) a
uniform magnetic flux is established, say 4>. How this is done (a current in a coil, or a permanent
magnet) is not important here.

In the center part of the window there is an iron cylinder (called rotor), free to rotate around its
axis. A coil of one turn is wound diametrically around the cylinder, parallel to its axis. As the
cylinder and its coil rotate, the flux through the coil changes. Figure 4.2 shows consecutive locations
of the rotor and we can see that the flux through the coil changes both in value and direction. The
top graph of figure 4.3 shows how the flux linkages of the coil through the coil would change, if the
rotor were to rotate at a constant angular velocity, uj.

A = l> cos [cut] (4.1)

47

48

CONCEPTS OF ELECTRICAL MACHINES; DC MOTORS

u

v y Y Y y Y Y

Fig. 4. 1 Geometry of an elementary DC motor

^y^^^yy M f T-Y J Hd^

3 4

Fig. 4.2 Flux through a coil of a rotating DC machine

Since the flux linking the coil changes with time, then a voltage will be induced in this coil, v,

Vcoil

dX
~dt

-\$u sin (ut)

coil->

(4.2)

shown in the second graph of figure 4.3. The points marked there correspond to the position of the
rotor in figure4.2.

This alternating voltage has to somehow be rectified, since this is a DC machine. Although this
can be done electronically, a very old mechanical method exists. The coil is connected not to the DC
source or load, but to two ring segments, solidly attached to it and the rotor, and hence rotating with
it. Two 'brushes', i.e. conducting pieces of material (often carbon/copper) are stationary and sliding
on these ring segments as shown in figure 4.4

The structure of the ring segments is called a commutator. As it rotates, the brushes make contact
with the opposite segments just as the induced voltage goes through zero and switches sign.

Figure 4.5 shows the induced voltage and the terminal voltage seen at the voltmeter of figure 4.4.
If a number of coils are placed on the rotor, as shown in figure 4.6, each connected to a commutator
segment, the total induced voltage to the coils, E will be:

E = k<\$>uo
where k is proportional to the number of coils.

(4.3)

GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS 49

Fig. 4.3 Flux and voltage in a coil of the DC machine in 4.2. Points 1-5 represent the coil positions.

Going back to equation 2.25,

k&uoi = Tuj
T = k\$i

(4.4)
(4.5)
(4.6)

If the electrical machine is connected to a load or a source as in figure4.7, the induced voltage and
terminal voltage will be related by:

vterm -^ ^q-^wdg

og-^wag for a generator

Vterm = E + i m R wdg for a motor

(4.7)
(4.8)

4.1.1 Example

A DC motor, when connected to a 100V source and to no load runs at 1200rpm. Its stator resistance
is 2Q. What should be the torque and current if it is fed from a 2201/ supply and its speed is
lbOOrpm? Assume that the field is constant.

The first piece of information gives us the constant k. Since at no load the torque is zero and
T = k&i = Ki, then the current is zero as well This means that for this operation:

V = E = k\$uj = Kcu

50 CONCEPTS OF ELECTRICAL MACHINES; DC MOTORS

Coil

Fig. 4.4 A coil of a DC motor and a commutator with brushes

100 200 300 400 500 600 700

Fig. 4.5 Induced voltage in a coil and terminal voltage in an elementary DC machine

but uo is 1200 rpm, or in SI units:

2tt
60 '

GEOMETRY, FIELDS, VOLTAGES, AND CURRENTS 51

And

Fig. 4.6 Coils on the rotor of DC machine

Fig. 4. 7 Circuit with a DC machine

100V = K • 125.66 => K = 0.796Vs
At the operating point of interest:

u = lbOOrpm = 1500— = lb7.08rad/s => E = Ku = 125V
60

For a motor:

V = E + IR

220 = 125 + / • 2ft

=> / = 47.5A

=> T = KI = 37.8lNm

52 CONCEPTS OF ELECTRICAL MACHINES; DC MOTORS

Notes

• The field of the DC motor can be created either by a DC current or a permanent magnet.

• These two fields, the one coming from the stator and the one coming from the moving rotor,
are both stationary (despite rotation) and perpendicular to each other.

• if the direction of current in the stator and in the rotor reverse together, torque will remain in
the same direction. Hence if the same current flows in both windings, it could be AC and the
motor will not reverse (e.g. hairdryers, power drills).

5^

Three-phase Windings

Understanding the geometry and operation of windings in AC machines is essential in understanding
how these machines operate. We introduce here the concept of Space Vectors, (or Space Phasors) in
its general form, and we see how they are applied to three-phase windings.

5.1 CURRENT SPACE VECTORS

Let us assume that in a uniformly permeable space we have placed three identical windings as shown
in figure 5.1. Each carries a time dependent current, i\(t), %2(t) and is(t). We require that:

h(t) + i 2 (t) + i 3 (t) = (5.1)

Each current produces a flux in the direction of the coil axis, and if we assume the magnetic
medium to be linear, we can find the total flux by adding the individual fluxes. This means that we
could produce the same flux by having only one coil, identical to the three, but placed in the direction
of the total flux, carrying an appropriate current. Figure 5.2 shows such a set of coils carrying for
i\ = 5 A, %2 = — 8 A and i 3 — = 3 A and the resultant coil.

To calculate the direction of the resultant one coil and the current it should carry, all we have to
do is create three vectors, each in the direction of one coil, and of amplitude equal to the current of
each coil. The sum of these vectors will give the direction of the total flux and hence of the one coil
that will replace the three. The amplitude of the vectors will be that of the current of each coil.

Let us assume that the coils are placed at angles 0°, 120° and 240°. Then their vectorial sum will
be:

i = i l + = i 1 +i 2 e> 12o0 +i 3 e> 24X > (5.2)

We call i, defined thus, a space vector, and we notice that if the currents ii, i 2 and i 3 are functions
of time, so will be the amplitude and the angle of i. By projecting the three constituting currents on
the horizontal and vertical axis, we can find the real (id = 5ft [i]) and imaginary (i q = 9[i]) parts of

53

54 THREE-PHASE WINDINGS

n

"wtht;

Fig. 5. 1 Three phase concentrated windings

(a)

(b)

(c)

Fig. 5.2 Currents in three windings (a), Resultant space vector (b), and corresponding winding position (c)

it. Also, from the definition of the current space vector we can reconstruct the constituent currents:

h

2 r

H(t)

f»[i(t)]

»[i(t)t

i 3 (t) = -»[i(t)e"^]

7 = 120° = — rad
' 3

(5.3)

(5.4)

STATOR WINDINGS AND RESULTING FLUX DENSITY 55

5.2 STATOR WINDINGS AND RESULTING FLUX DENSITY

Fig. 5.3 A Stator Lamination

Stator Winding

Stator

Airgap

Rotor

Fig. 5.4 A sinusoidal winding on the stator

Assume now that these windings are placed in a fixed structure, the stator, which is surrounds a
rotor. Figure 5.3 shows a typical stator cross-section, but for the present we'll consider the stator
as a steel tube. Figure 5.5 shows the windings in such a case. Instead of being concentrated, they
are sinusoidally distributed as shown in figure 5.4. Sinusoidal distribution means that the number of

56 THREE-PHASE WINDINGS

turns dN s covering an angle dO at a position 6 and divided by dO is a sinusoidal function of the angle
6. This turns density, n s i(9), is then:

dn s
~dB

n s \{8) = n s sin 6

and for a total number of turns N s in the winding:

N s = n sl (0)dO =» n<
Jo

AT

lW = _i sin(

We now assign to the winding we are discussing a current i\. To find the flux density in the airgap

Current in
positive direction

Stator

Current in
negative direction

Rotor

Stator winding

Fig. 5.5 Integration path to calculate flux density in the airgap

between rotor and stator we choose an integration path as shown in figure 5.5. This path is defined by
the angle 6 and we can notice that because of symmetry the flux density at the two airgap segments
in the path is the same. If we assume the permeability of iron to be infinite, then H iron = and:

2H gl (0)g

2ggi(g)

Mo

Bgl(0)

r9 + >K

\ iin 8 i(</>)d<l>
Jo

i\N s cos 6

i\ cosf

1g

(5.5)

This means that for a given current i\ in the coil, the flux density in the air gap varies sinusoidally with
angle, but as shown in figure 5.6 it reaches a maximum at angle 6 = 0. For the same machine and
conditions as in 5.6, 5.7 shows the plots of turns density, n s (6) and flux density, B g (Q) in cartesian
coordinates with in the horizontal axis.

If the current i\ were to vary sinusoidally in time, the flux density would also change in time,
maintaining its space profile but changing only in amplitude. This would be considered a wave, as it

STATOR WINDINGS AND RESULTING FLUX DENSITY 57

Fig. 5.6 Sketch of the flux in the airgap

Fig. 5.7 Turns density on the stator and air gap flux density vs.

changes in time and space. The nodes of the wave, where the flux density is zero, will remain at 90°
and 270°, while the extrema of the flux will remain at 0° and 180°.

Consider now an additional winding, identical to the first, but rotated with respect to it by 120°.
For a current in this winding we'll get a similar airgap flux density as before, but with nodes at
90° + 120° and at 270° + 120°. If a current i% is flowing in this winding, then the airgap flux density

58 THREE-PHASE WINDINGS

due to it will follow a form similar to equation 5.5 but rotated 120° = ^f.

B g2 (0)=i 2 I ^co S (e-f) (5.6)

Similarly, a third winding, rotated yet another 120° and carrying current i 3 , will produce airgap
flux density:

B g3 (0) = i 3 ^co S (e-f) (5.7)

Superimposing these three flux densities, we get yet another sinusoidally distributed airgap flux
density, that could equivalently come from a winding placed at an angle <fi and carrying current i:

B g {0) = B gl (0) + B g2 (6) + B g3 (6) = i^ cos((9 + </>) (5.8)

This means that as the currents change, the flux could be due instead to only one sinusoidally
distributed winding with the same number of turns. The location, (j){t), and current, i(t), of this
winding can be determined from the current space vector:

i(t) = i(*) Z *« = h{t) + t 2 (*)e>' 120 ° + i 3 (t)e^ °

5.2.1 Balanced, Symmetric Three-phase Currents

If the currents ii, i 2 , is form a balanced three-phase system of frequency f s = uj s /2tt, then we can
write:

\/2
h = V2Icos(u s t + 0i) = -y [l a eP Uat + I s e- J " st ]

i 2 = V2IcO8(c0 s t -0! + —) = — [l se i(-^-2-/3) + Ifle -jW-27r/3)l (5 9)

O Z L

i 3 = v / 2/co S Kt-0 1 + ^) = ^[l se ^^ t - 4 -/ 3 )+I se -^^ t - 4 -/ 3 )

O Z -

where I is the phasor corresponding to the current in phase 1. The resultant space vector is

i i (t) = |^I e i«.* = |^j e J(".*+*i) I = JeWi+f) (5.10)

The resulting flux density wave is then:

B{0,t) = ? ) V2I I ^cos(uj s t + 4> 1 -0) (5.11)

z zg

which shows a travelling wave, with a maximum value 5 = |\/2^— • This wave travels around the
stator at a constant speed u s , as shown in figure 5.8

5.3 PHASORS AND SPACE VECTORS

It is easy at this point to confuse space vectors and phasors. A current phasor, I = Ie^°, describes
one sinusoidally varying current, of frequency u, amplitude y/2I and initial phase 0q. We can

PHASORS AND SPACE VECTORS 59

Fig. 5.8 Airgap flux density profile, £3 > £2 > ti

reconstruct the sinusoid from the phasor:

i(t)

V2

[le jUJt + Te~ juJt ] = V2Icos(u;t + O ) = ^ {^ ju

(5.12)

Although rotation is implicit in the definition of the phasor, no rotation is described by it.

On the other hand, the definition of a current space vector requires three currents that sum to
zero. These currents are implicitly in windings symmetrically placed, but the currents themselves
are not necessarily sinusoidal. Generally the amplitude and angle of the space vector changes with
time, but no specific pattern is a priori defined. We can reconstruct the three currents that constitute
the space vector from equation 5.3.

When these constituent currents form a balanced, symmetric system, of frequency uj s , then the
resultant space vector is of constant amplitude, rotating at constant speed. In that case, the relationship
between the phasor of one current and the space vector is shown in equation 5.10.

5.3.1 Example

Let us take three balanced sinusoidal currents with amplitude 1, i.e. rms value of l/y/2 A. Choose
an initial phase angle such that:

When out = 0, as shown in figure 5.9a,

h(t)

=

1 cos{ujt)A

*2(*)

=

1 cos(t^ - 2tt/3) A

i 2 (t)

=

1 cos(t^ - 4tt/3) A

5.9 a,

h

= \A

22

= -0.5 A

^3

= -0.5 A

60 THREE-PHASE WINDINGS

and later, when out = 20° = 7r/9 rad, as shown in figure 5.9b,

h = 0.939A
i 2 = -0.766A
i 3 = -0.174A

/i 3 =-0.5A

/

i =1A

i

\ i 2 = -0.5A

(a)

(b)

Fig. 5.9 Space vector movement for sinusoidal, symmetric three-phase currents

5.4 MAGNETIZING CURRENT, FLUX AND VOLTAGE

Let us now see what results this rotating flux has on the windings, using Faraday's law. From this
point on we'll use sinusoidal symmetric three-phase quantities.

We look again at our three real stationary windings linked by a rotating flux. For example, when
the current is maximum in phase 1, the flux is as shown in figure 5.10a, linking all of the turns in
phase 1. Later, the flux has rotated as shown in figure 5.10b, resulting in lower flux linkages with
the phase 1 windings. When the flux has rotated 90°, as in 5.10c the flux linkages with the phase 1
winding are zero.

To calculate the flux linkages A we have to take a turn of the winding, placed at angle 0, as shown
in figure 5.11. The flux through this coil is:

r9 r9+

\$(£, 6) = / B g (t, <j>)dA = lr B {

9 (t,

(5.13)

But the number of turns linked by this flux is dn s (0) = n s (6)d6, so the flux linkages for these turns
are:

d\ = n s (0)d0 .\$(0)

MAGNETIZING CURRENT, FLUX AND VOLTAGE 61

(a)

(b)

(c)

Fig. 5.10 Rotating flux and flux linkages with phase 1

Fig. 5.11 Flux linkages of one turn

To find the flux linkages Ai for all of the coil, we have to integrate the flux linkages over all turns of
coil 1:

Ai

X(0)dO

giving us at the end:

AiW

N 2 s lr

37r/iov2/cos(a;t + <fii) = Lm \2 1 cos (cut + <pi)

(5.14)

which means that the flux linkages in coil 1 are in phase with the current in this coil and proportional
to it. The flux linkages of the other two coils, 2 and 3, are identical to that of coil 1, and lagging in
time by 120° and 240°. With these three quantities we can create a flux-linkage space vector, A.

\ — \ _i_ \ J 120° i \ ^'240° t •

A = Ai + A2e J + Aze J = Lm i

(5.15)

Since the flux linkages of each coil vary, and in our case sinusoidally, a voltage is induced in each
of these coils. The induced voltage in each coil is 90° ahead of the current in it, bringing to mind

62

THREE-PHASE WINDINGS

the relationship of current and voltage of an inductor. Notice though, that it is not just the current in
the winding that causes the flux linkages and the induced voltages, but rather the current in all three
windings. Still, we call the constant Lm magnetizing inductance.

ei(t) =

CLA\

e 2 {t) =

d\ 2

e 3 (t) =

d\ 3
dt

■ coV2I cos(ut -

■ coV2I cos(cot -

■ UJ

V2I cos {uot -

and of course we can define voltage space vectors e:

ei H- e 2 e

J 120°

e 3 e-

j240 u

7T 27I\

2 3 }

7T 47T N

jcjL m i

(5.16)

(5.17)

Note that the flux linkage space vector A is aligned with the current space vector, while the voltage
space vector e is ahead of both by 90°. This agrees with the fact that the individual phase voltages
lead the currents by 90°, as shown in figure 5.12.

Fig. 5. 12 Magnetizing current, flux-linkage and induced voltage space vectors

6

Induction Machines

Induction machines are often described as the 'workhorse of industry'. This dice reflects the reality
of the qualities of these machines. They are cheap to manufacture, rugged and reliable and find their
way in most possible applications. Variable speed drives require inexpensive power electronics and
computer hardware, and allowed induction machines to become more versatile. In particular, vector
or field-oriented control allows induction motors to replace DC motors in many applications

6.1 DESCRIPTION

The stator of an induction machine is a typical three-phase one, as described in the previous chapter.
The rotor can be one of two major types. Either a) it is wound in a fashion similar to that of the stator
with the terminals led to slip rings on the shaft, as shown in figure 6.1, or b) it is made with shorted

Slip
Rings

Shaft

Rotor

Fig. 6. 1 Wound rotor slip rings and connections

bars. Figure 6.2 shows the rotor of such a machine, while figures 6.3 show the shorted bars and the
laminations.

The picture of the rotor bars is not easy to obtain, since the bars are formed by casting aluminum
in the openings of the rotor laminations. In this case the iron laminations were chemically removed.

63

64 INDUCTION MACHINES

Fig. 6.2 Rotor for squirrel cage induction motor

(a) Rotor bars and rings

(b) Rotor

Fig. 6.3 Rotor Components of a Squirrel Cage Induction Motor

6.2 CONCEPT OF OPERATION

As these rotor windings or bars rotate within the magnetic field created by the stator magnetizing
currents, voltages are induced in them. If the rotor were to stand still, then the induced voltages
would be very similar to those induced in the stator windings. In the case of squirrel cage rotor, the
voltage induced in the bars will be slightly out of phase with the voltage in the next one, since the
flux linkages will change in it after a short delay.

If the rotor is moving at synchronous speed, together with the field, no voltage will be induced in
the bars or the windings.

CONCEPT OF OPERATION 65

13

(a) Rotor bars in the stator field

(b) Voltages in rotor bars

Fig. 6.4

Generally when the synchronous speed is uj s = 27r/ s , and the rotor speed cj , the frequency of
the induced voltages will be f r , where 2nf r = uj s — ujq. Maxwell's equation becomes here:

8 = v x B g
where v is the relative velocity of the rotor with respect to the field:

v = (cj s - LUo)r

(6.1)

(6.2)

Since a voltage is induced in the bars, and these are short-circuited, currents will flow in them. The
current density J (0) will be:

~J(0) = -~S (6.3)

P

These currents are out of phase in different bars, just like the induced voltages. To simplify the
analysis we can consider the rotor as one winding carrying currents sinusoidally distributed in space.
This will be clearly the case for a wound rotor. It will also be the case for uniformly distributed rotor
bars, but now each bar, located at an angle will carry different current, as shown in figure 6.5 a:

J = -(uj a -coo) • B g (#)

J(0) = -(u a -u )B g sm(0)

We can replace the bars with a conductive cylinder as shown in figure 6.5 b.
We define as slip s the ratio:

LO s - UJq

(6.4)
(6.5)

(6.6)

66

INDUCTION MACHINES

Equivalent

Conducting

Sheet

(a) Currents in rotor bars

(b) Equivalent Current Sheet in Rotor

Fig. 6.5 Current Distribution in equivalent conducting sheet

At starting the speed is zero, hence 5 = 1, and at synchronous speed, uu s = uuq, hence 5 = 0. Above
synchronous speed s < 0, and when the rotor rotates in a direction opposite of the magnetic field

5>1.

6.2.1 Example

The rotor of a two-pole 3 -phase induction machine rotates at 3300rpm, while the stator is fed by a
three-phase system of voltages, at 60Hz. What are the possible frequencies of the rotor voltages?
At 3300 rpm

uj = 3300 — - = 34b.6rad/s while uu s = 377rad/s

These two speeds can be in the opposite or the same direction, hence:

uo r = uo s — uj = 377 =b 345.6 = 722.bSrad/s or 31A3rad/s

f r = llbHz or = bHz

6.3 TORQUE DEVELOPMENT

We can now calculate forces and torque on the rotor. We'll use the formulae:

F = Bli, T = F-r

(6.7)

since the flux density is perpendicular to the current. As / we'll use the length of the conductor, i.e.
the depth of the motor. We consider an equivalent thickness of the conducting sheet d e . A is the
cross section of all the bars.

J\ — Tlrotor bars . & — Z7TVCl e
Tl ro f or bars®

8r

(6.8)
(6.9)

OPERATION OF THE INDUCTION MACHINE NEAR SYNCHRONOUS SPEED 67

(a)

(b)

Fig. 6.6 Calculation of Torque

For a small angle dO at an angle 0, we calculate the contribution to the total force and torque:

dF
dF
dT

(JdA) -Bg-l, B g = B g sin(0)

(Jd e r d0)B g

rdF

'2nr 2 ld t

/ dT

Je=o

B Au) 8 -u) )

(6.10)

Flux density in the airgap is not an easy quantity to work with, so we can use the relationship
between flux density (or flux linkages) and rotor voltage and finally get:

8 rd e

A 2 s (u s -u )

where

A,

(6.11)

Although the constants in equations 6.10 and 6.11 are important we should focus more on the
variables. We notice that in equation 6.10 the torque is proportional to the frequency of the rotor
currents, (u s — coq) and the square of the flux density. This is so since the torque comes from the
interaction of the flux density B g and the rotor currents. But the rotor currents are induced (induction
motor) due to the flux B g and the relative speed uo s — uoq.

On the other hand, equation 6.1 1 gives us torque as a function of more accessible quantities, stator
induced voltage E s and frequency uo s . This is so, since there is a very simple and direct relationship
between stator induced voltage, flux (or flux linkages) and frequency.

6.4 OPERATION OF THE INDUCTION MACHINE NEAR SYNCHRONOUS SPEED

We already determined that the voltages induced in the rotor bars are of slip frequency, f r =
(uj s — luo)/(2it). At rotor speeds near synchronous, this frequency, f r is quite small. The rotor bars
in a squirrel cage machine possess resistance and leakage inductance, but at very low frequencies,

68

INDUCTION MACHINES

i.e near synchronous speed, we can neglect this inductance. The rotor currents therefore are limited
near synchronous speed by the rotor resistance only.

The induced rotor-bar voltages and currents form space vectors. These are perpendicular to the
stator magnetizing current and in phase with the space vectors of the voltages induced in the stator
as shown in figure 6.7 and figure

Fig. 6. 7 Stator Magnetizing Current, airgap flux and rotor currents

These rotor currents, i r produce additional airgap flux, which is 90° out of phase of the magnetizing
flux. But the stator voltage, e s , is applied externally and it is proportional to and 90° out of phase of
the airgap flux. Additional currents, i sr will flow in the stator windings in order to cancel the flux
due to the rotor currents. These currents are shown in figures 6.8. In 6.9 the corresponding space
vectors are shown.

(a) Rotor Current and Stator Current Components

(b) Space Vectors of the Rotor and Stator Cur-
rents and induced voltages

Fig. 6.8 Rotor and Stator Currents in an Induction Motor

OPERATION OF THE INDUCTION MACHINE NEAR SYNCHRONOUS SPEED 69

'SFR

^ sm

Fig. 6.9 Space- Vectors of the Stator and Rotor Current and Induced Voltages

There are a few things we should observe here:

• i sr is 90° ahead of i sm , the stator magnetizing current. This means that it corresponds to
currents in windings i\ r , %2 r , ^3r, leading by 90° the magnetizing currents ii m , i2m, ^3m-

• The amplitude of the magnetizing component of the stator current is proportional to the stator
frequency, f s and induced voltage. On the other hand, the amplitude of this component of the
stator currents, i sr , is proportional to the current in the rotor, i r , which in turn is proportional
to the flux and the slip speed, uj r = uj s — ujo, or proportional to the developed torque.

• We can, therefore split the stator current of one phase, i s i, into two components: One in phase
with the voltage, i sr i and one 90° behind it, i 8m \. The first reflects the rotor current, while
the second depends on the voltage and frequency. In an equivalent circuit, this means that i sr \
will flow through a resistor, and i sm i will flow through an inductor.

• Since i sr i is equal to the rotor current (through a factor), it will be inversely proportional to
ou s — cj r , or, better, proportional to uu s /(uu s — uj r ). Figure 6.10 reflects these considerations.

'S1

'sm

"sr

k m

CO,

CO;

COq

Fig. 6.10 Equivalent circuit of one stator phase

If we supply our induction motor with a three-phase, balanced sinusoidal voltage, we expect that
the rotor will develop a torque according to equation 6.11. The relationship between speed, t^o and
torque around synchronous speed is shown in figure 6.1 1. This curve is accurate as long as the speed
does not vary more than ±5% around the rated synchronous speed uj s .

70 INDUCTION MACHINES

Fig. 6.11 Torque-speed characteristic near synchronous speed

We notice in 6.1 1 that when the speed exceeds synchronous, the torque produced by the machine
is of opposite direction than the speed, i.e. the machine operates as a generator, developing a torque
opposite to the rotation (counter torque) and transferring power from the shaft to the electrical system.

We already know the relationship of the magnetizing current, I sm to the induced voltage E srn
through our analysis of the three-phase windings. Let us now relate the currents i r and i sr with the
same induced voltage.

The current density on the rotor conducting sheet J is related to the value of the airgap flux
density B g through:

(6.12)

J = -{u 8 -u )B t
P

This current density corresponds to a space vector i r that is opposite to the i sr in the stator. This
current space vector will correspond to the same current density:

J

rN*

1

(6.13)

while the stator voltage e s is also related to the flux density B g . Its amplitude is:

: U s

-NJrB n

(6.14)

Finally, substituting into 6.12, and relating phasors instead of space vectors, we obtain:

(J.

V s =Rr-

Is

(6.15)

Using this formulation we arrive at the formula for the torque:

Ei 1 us

■wo

U s Rr

o A '

Mr

3P n

(6.16)

where A = (E s /u s )) Here P g is the power transferred to the resistance Rr-

g io tn^ F uww uanoiuiiuu iu tn^ n.oi^anvu it^ - _ u , through the
airgap. Of this power a portion is converted to mechanical power represented by losses on resistance
Rr - ^_° , and the remaining is losses in the rotor resistance, represented by the losses on resistance
Rr. Figure 6.12 shows this split in the equivalent circuit. Note that the resistance Rr - ^_° can be
negative, indicating that mechanical power is absorbed in the induction machine.

6.4.1 Example

A 2-pole three-phase induction motor is connected in Y and is fed from a 60 Hz, 208 V (/ — I) system.
Its equivalent one-phase rotor resistance is Rr = 0.112511 At what speed and slip is the developed
torque 2SNm?

■s1

■sr

LEAKAGE INDUCTANCES AND THEIR EFFECTS 71

~\

"sm

k m

R R ^n

> R R

co c

CO

s^

CO;

-%

J

Fig. 6. 12 Equivalent circuit of one stator phase separating the loss and torque rotor components

Is

v\A-^mnr^-

-OTT^-

■\

■s1

p sm

c s

'm

"sr

R R M n

V R R

C0 C

(0;

■°\)

y

Fig. 6.13 Complete equivalent circuit of one stator phase

T

28
s

LJ

120

Rr

-uj r with V, = 1201/

1

,377/ 0.1125
uu r 10.364

0.0275

377

6.5 LEAKAGE INDUCTANCES AND THEIR EFFECTS

In the previous discussion we assumed that all the flux crosses the airgap and links both the stator
and rotor windings. In addition to this flux there are flux components which link only the stator or
the rotor windings and are proportional to the currents there, producing voltages in these windings
90° ahead of the stator and rotor currents and proportional to the amplitude of these currents and
their frequency.

This is simple to model for the stator windings, since the equivalent circuit we are using is of the
stator, and we can model the effects of this flux with only an inductance. The rotor leakage flux can
be modelled in the rotor circuit with an inductance Li s , as well, but corresponding to frequency of
fr = UJs 2t T UJ ° > tne f rec l uenc y °f the rotor currents. It turns out that when we try to see its effects on
the stator we can model it with an inductance L ir at frequency f s , as shown in the complete 1 -phase
equivalent circuit in figure 6.13.

72 INDUCTION MACHINES

Here E s is the phasor of the voltage induced into the rotor windings from the airgap flux, while
V s is the phasor of the applied 1 -phase stator voltage. The torque equations discussed earlier, 6.16,
still hold, but give us slightly different results: We can develop torque-speed curves, by selecting
speeds, solving the equivalent circuit, calculating power P g , and using equation 6.16 for the torque.
Figure 6.14 shows these characteristics for a wide range of speeds.

Fig. 6. 14 Torque, current and power factor of an induction motor vs. speed

6.6 OPERATING CHARACTERISTICS

Figure 6.14 shows the developed torque, current, and power factor of an induction motor over a
speed range from below zero (slip > 1) to above synchronous (slip < 0). It is clear that there are
three areas of interest:

1 . For speed < uu < uu s the torque is of the same sign as the speed, and the machine operates
as a motor. There are a few interesting point on this curve, and on the corresponding current
andp.f. curves.

2. For speed ujq < 0, torque and speed have opposite signs, and the machine is in breaking mode.
Notice that the current is very high, resulting in high winding losses.

3. for speed uj > uj s the speed and torque are of opposite signs, the machine is in generating
mode, and the current amplitude is reasonable.

Let us concentrate now on the region < uj < uj s . The machine is often designed to operate as
a motor, and the operating point is near or exactly where the power factor is maximized. It is for this

OPERATING CHARACTERISTICS 73

R Th X Th X lr

w-^nnp nnn^ — , ^

R

L

l R

V T h 'sr ( V R R M s

R D _^L_ ( W ^

Fig. 6. 15 Equivalent circuit of the stator with Thevenin equivalent of the stator components

point that the motor characteristics are given on the nameplate, rated speed, current, power factor
and torque. When designing an application it is this point that we have to consider primarily: Will
the torque suffice, will the efficiency and power factor be acceptable?

A second point of interest is starting, ( slip 5 = 1) where the torque is not necessarily high,
but the current often is. When selecting a motor for an application, we have to make sure that this
starting torque is adequate to overcome the load torque, which may also include a static component. In
addition, the starting current is often 3-5 times the rated current of the machine. If the developed torque
at starting is not adequately higher than the load starting torque, their difference, the accelerating
torque will be small and it may take too long to reach the operating point. This means that the current
will remain high for a long time, and fuses or circuit breakers may operate.

A third point of interest is the maximum torque, T max , corresponding to speed uiTmax- We can
find it by analytically calculating torque as a function of slip, and equating the derivative to 0. This
point is interesting, since points to the right of it correspond in general to stable operating conditions,
while point to its left correspond to unstable operating conditions.

We can study this point if we take the Thevenin equivalent circuit of the left part of the stator
equivalent circuit, including, V 39 R s , X[ s and X m . This will give us the circuit in figure 6.15.

Using the formula 6.16 we arrive at:

vk {**)

(RTh + E f) +(X Th + X lr )

2

(6.17)

The maximum torque will develop when the airgap power, P g , i.e. the power delivered to Rr/s,
is maximum, since the torque is proportional to it. Taking derivative of 6.17, we find that maximum
torque will occur when:

Rr

SmaxT

(R Th ) 2 + (X Th + X lr ) 2 or (6.18)

SmaxT = , (6.19)

^Th + (^Th + Xi r )

giving maximum torque:

T max = 3^ = Vth == (6.20)

S Rrh + y RtH + {^Th + Xi r )

74

INDUCTION MACHINES

Fig. 6.16 Effect of changing rotor resistance on the torque-speed and current speed characteristic

If we neglect the stator resistance we can easily show that the general formula for the torque
becomes:

- (6.21)

+

If we neglect both the stator resistance and the magnetizing inductance, we can develop simple
formulae for T max and uoTmax- To do so we have to assume operation near synchronous speed,
where that value of Rr — ^ — is much larger than uu s Li r .

LO g LO r\

^Tmax

T

UJ S

3
2

Rr

Li r + Lis
2 -,

Ri

-(U 8

■ Mm

Lis + Li r

(6.22)
(6.23)

We notice here that the slip frequency at this torque, uj r = co s — ^Tmax-> fc> r a constant flux
A s = ^f- is independent of frequency and proportional to the resistance Rr. We already know
that this resistance is proportional to the rotor resistance, so if the rotor resistance is increased, the
torque-speed characteristic is shifted to the left, as shown in figure 6.16.

If we have convenient ways to increase the rotor resistance, we can increase the starting torque,
while decreasing the starting current. Increasing the rotor resistance can be easily accomplished in a
wound-rotor machine, and more complex in squirrel cage motor, by using double or deep rotor bars.

In the formulae developed we notice that the maximum torque is a function of the flux. This
means that we can change the frequency of the stator voltage, but as long as the voltage amplitude
changes so that the flux stays the same, the maximum torque will also stay the same. Figure 6.17
shows this. This is called Constant Volts per Hertz Operation and it is a first approach to controlling
the speed of the motor through its supply.

STARTING OF INDUCTION MOTORS 75

f =20Hz

s \

f =35Hz\

S V.

f =50Hz^

s

\ \

Fig. 6.17 Effect on the Torque - Speed characteristic of changing frequency while keeping flux constant

Near synchronous speed the effect of the rotor leakage inductance can be neglected, as discussed
earlier. This assumption gives us the approximate torque-speed equation 6.16 discussed earlier.

El 1 u a

UJQ

AtLJr

3P

Ld s Rr LO s Rr LO s

Figure 6.18 shows both exact and approximate torque-speed characteristics. It is important to notice
that the torque calculated from the approximate equation is grossly incorrect away from synchronous
speed.

6.7 STARTING OF INDUCTION MOTORS

To avoid the problems associated with starting (too high current, too low torque), a variety of
techniques are available.

An easy way to decrease the starting current is to decrease the stator terminal voltage. One can
notice that while the stator current is proportional to the in voltage, torque will be proportional to
its square. If a transformer is used to accomplish this, both developed torque and line current will
decrease by the square of the turns ratio of the transformer.

A commonly used method is to use a motor designed to operate with the stator windings connected
in A, and have it connected in Y at starting. As the voltage ratio is.

V a ,Y = 4=V»,A (6.24)

Vs

then

i

U,Y

t s ,A

(6.25)

(6.26)

76 INDUCTION MACHINES

\. \

\ 4

\ t

\ I

\l
\l

maxT

V w
1 s

y

y

y

y

y

F/g. 5. 78 Exact and approximate torque-speed characteristics

1 line.Y

1 MneA

Fig. 6. 19 Y- Delta starting of an induction motor

But in a A connection, Iu ne = VSI p h, leading to:

1
3 J

(6.27)

Once the machine has approached the desired operating point, we can reconfigure the connection
to A, and provide better efficiency.

This decrease in current is often adequate to allow a motor to start at low load starting torque.
Using a variable frequency and voltage supply we can comfortably increase the starting torque, as
shown in figure 6.17, while decreasing the starting current.

6.8 MULTIPLE POLE PAIRS

If we consider that an induction machine will operate close to synchronous speed (3000rpm for 50Hz
and 3600rpm for 60 Hz) we may find that the speed of the machine is too high for an application.
If we recall the pictures of the flux in AC machines we have seen, we can notice that the flux has a
relatively long path to travel in the stator making the stator heavy and lossy.

MULTIPLE POLE PAIRS 77

Fig. 6.20 Equivalent windings for a 6 pole induction motor

A machine with more than one pole pair is quite similar to that with only one. The difference is
that for example in a 4 pole machine each side of a sinusoidally distributed winding of one phase
covers only 90° instead of 180°. A result is that there is room for four rather than two coil sides of
each phase. Figure 6.20 shows at one instant the equivalent windings resulting from the the three
phase windings.

The effects of a large number of poles on the operation of the machine are easy to predict. If the
machine has p poles, or p/2 pairs of poles, in one period of the voltage the flux will travel -w s rad/s.
Hence the rotor speed corresponding to synchronous will be uu sm :

2

uj sm = -uj s (6.28)

P

We introduce now the actual, mechanical speed of the rotor, cc; m , while we keep the term uu as
the rotor speed of a two pole motor. We generally measure uum in rad/s, while we measure c^o in
electrical rad/s. We retain the same definition for slip based on the electrical speed c^o .

2
uj m = -uj (6.29)

P

UJ S -UJ Us- l^ra

s = = (6.30)

This means that for a 4 pole machine, supplied from a source of at 60Hz, and operating close to
rated conditions, the speed will be near 1800rpm, while for a 6 pole machine, the speed will be near
1200rpra.

While increasing the number of poles results in a decrease of the synchronous and operating
speeds of the machine, it also results in an increase of the developed torque of the machine by the
same ratio. Hence, the corrected torque formula will be

T = 3 l^ = 3 P^l (6 . 31)

2 uj s 2wq

Similarly, the torque near the synchronous speed is:

pEl 1 uj s -ujq = pA^. =3 P^o_
2 uj s R R uj s 2 R R 2 uj s

78 INDUCTION MACHINES

while the previously developed formulas for maximum torque will become:

2 2c0s R Th + Jr 2 TIi + (X T k + x lr y

(6.32)

and

3p/T4\ 2 1 3p/V.\ 2 1

T — -22UJ ^ K " WTmJ -22UJ ^T^ ( }

6.8.1 Example

A 3-phase 2-pole induction motor is rated 190V, 60Hz, it is connected in Y, and has R r = 6.6Q,
R s = 3.1^, Xm = 190f2, Xi r = 100, and X i s = 311 Calculate the motor starting torque, starting
current and starting power factor under rated voltage. What will be the current and power factor if
no load is connected to the shaft?

1. At starting s = 1:

190 .„, ,.,__ --,Z-64.5°

I s = /{[3.1+j3]+il90||(6.6+j'10)} = 7.06^- 54 - 5 A

I R = I s £59 = 6.7^™° A

6.6 + jl0 + jl90

T = 3^ = 3^-^ 2 =2.36iVrn
u 8 2 377 2

2. Under no load the speed is synchronous and s = 0:

I s = 110/ [3.1 + jS + J190] = 0.57 Z " 891 ° A
I 8 = 0.57A
pf = O.OWlagging

6.8.2 Example

A 3-phase 2-pole induction motor is rated 190V, 60Hz it is connected in Y, and has R r = 6.6fl,
R s = 3.m, Xm = 190O, X\ T = 10O, and X\ s = 3Q. It is operating from a variable speed -
variable frequency source at a speed of 1910rpm, under a constant V/f policy and the developed
torque is O.SNm. What is the voltage and frequency of the source? (Hint: Calculate first the slip).
The ratio V s /co s stays 110/377.

T =

PnfVsV 1

0.8 =

= 1 • 3 I —— J — -u r ^u r = 20.65 rad/s

U s =

= u m %+LJ R = 220.66

2 5

■/- =

= 35#z => V s = 220.66^ = 64.4V or IIOV1-1

377

6.8.3 Example

A 3-phase 4-pole induction machine is rated 230V, 60Hz. It is connected in Y and it has R r =
0.191^, R s = 0.2^, Lm = 35mH, L\ r = l.bmH, andL\ s = 1.2mH. It is operated as a generator

MULTIPLE POLE PAIRS 79

connected to a variable frequency /variable voltage source. Its speed is 2036rpm, with counter- torque
of59Nm. What is the efficiency of this generator? (Hint: here power in is mechanical, power out
is electrical; calculate first the slip)

Although we do not know the voltage or the frequency, we know their ratio since it is always

132.8/377.

2 \lJsJ Rr

r . , 132. 8\ 2 1
-59 = 3 • 2

377 ) 0.191
=4> uj r = —15.14 rad/s

Now we can find the synchronous speed, by adding slip and rotor speeds:

p 2tt • 2036

u s = oo m - + u r = — 2 — 15.14 = 411.3 rad/s

Zj ou

132

=>f 3 = 65.5#z => V s = 65.5 = 144V

60

We have to recalculate the impedances of the equivalent circuit for the frequency of 65.5Hz:
X m = 35 • 10" 3 • 411.3 = 14.41], X u = 0.49O, X lr = 0.617ft

R R - ' m2 = -5.38ft

I s = 144/ [0.2 + jO.49 + jl4.4| |(0.191 - 5.38 + J0.617)] = 30 Z " 148 ° A

I R = 27.2 Z - 166 - 9 A

Notice that with generation operation Rr < 0. We can calculate now losses etc.

P m = 3 • 27.2 2 5.38 = 11.941W

rotordoss

Protor,loss = 3 • 27.2 2 0.191 = 423W

Pstator,loss = 3 • 30 2 0.2 = 540W

= ^ > *out -* m -Lrotor,loss ^stator,loss lU.yoUrvVT

p

^r] = -^=0.919

7

Synchronous Machines and

Drives

We noticed in discussing induction machines that as the rotor approaches synchronous speed, the
frequency of the currents in the rotor decreases, as does the amplitude of these currents. The reason
an induction motor produces no torque at synchronous speed is not that the currents are DC, but
rather that their amplitude is zero.

It is possible to operate a three-phase machine at synchronous speed if DC is externally applied
to the rotor and the rotor is rotated at synchronous speed. In this case torque will be developed only
at this speed, i.e. if the rotor is rotated at speeds other than synchronous, the average torque will be
zero.

Machines operating on this principle are called synchronous machines, and cover a great variety.
As generators they can be quite large, rated a few hundred MVA, and almost all power generation
is through these machines. Large synchronous motors are not very common, but can be an attractive
alternative to induction machines. Small synchronous motors with permanent magnets in the rotor,
rather than coils with DC, are rapidly replacing induction motors in automotive, industrial and
residential applications, since they are more efficient and lighter.

7.1 DESIGN AND PRINCIPLE OF OPERATION

The stator of a synchronous machine is of the type that we have already discussed, with three
windings carrying a three-phase system of currents. The rotor can be one of two distinct types:

7.1.1 Wound Rotor Carrying DC

In this case the rotor steel structure can be either cylindrical, like that in figure 7. la, or salient like the
one in 7.1b. In either of these cases the rotor winding carries DC, provided to it through slip rings, or
through a rectified voltage of an inside-out synchronous generator mounted on the same shaft. Here
we'll limit ourselves to discussing only cylindrical rotors.

81

82

SYNCHRONOUS MACHINES AND DRIVES

Sinusoidally

Distributed

Winding

Concentrated
Winding

(a) Cylindrical rotor

(b) Salient pole rotor

Fig. 7.1

7.1.2 Permanent Magnet Rotor

In this case instead of supplying DC to the rotor we create a magnetic field attached to it by adding
magnets on the rotor. There are many ways to do this, as shown in figure 7.2, and all have the
following effects:

• The rotor flux can no longer be controlled externally. It is defined uniquely by the magnets
and the geometry,

• The machine becomes simpler to construct, at least for small sizes.

7.2 EQUIVALENT CIRCUIT

The flux in the air gap can be considered to be due to two sources: the stator currents, and the rotor
currents or permanent magnet. We have discussed already how the currents in the stator produce flux.
Remember that this flux could also be produced by one equivalent winding, rotating at synchronous
speed and carrying current equal to the magnitude of the stator-current space vector.

The rotor is itself such a winding, a real one, sinusoidally distributed, carrying DC and rotating at
synchronous speed. It produces an airgap flux, which could also be produced by an additional set of
three phase stator currents, giving a space vector i R . The amplitude of this space vector would be:

M = £«, <7.D

where N s is the number of the stator turns of the one equivalent winding and Nr is the number of
the turns in the rotor winding. Its angle <j) R would be the same as the angle of the rotor position:

LU,t-

'JRO

(7.2)

The stator current space vector has amplitude:

= -V2I s
2 s

(7.3)

EQUIVALENT CIRCUIT 83

(i) Surface PM (SPM) synchronous
machine

(ii) Surface inset PM (SIPM)
synchronous machine

(iii) Interior PM (IPM) synchronous

machine

(iv) Interior PM with circumferential
orientation synchronous machine

Fig. 7.2 Possible magnet placements in PMAC motors

where I s is the rms current of one phase. The stator current space vector will have an instantaneous
angle,

(j) is = uj s t + (f) is o (7.4)

The airgap flux then is produced by both these current space vectors (rotor and stator). This flux
induces in the stator windings a voltage, e s . In quasi steady-state everything is sinusoidal and the
voltage space vector corresponds to three phase voltages Ei, E 2 , E 3 . In this case we can create an
equivalent circuit for the stator, 7.3. Here If is the stator AC current, that if it were to flow in the
stator windings would have the same effects as the rotor current, if. In our analysis we can use as
reference either the stator voltage, V s , or the stator current, I s . Figure 7.4. There are some angles
to notice in this figure. We call 6 the power factor angle, i.e. the angle between I s and V s . We call
(3, the angle between V s and If, and power angle, S, that between If and Is.

A few relationships to notice here:

• The space vector of the voltages induced in the stator, e s , is 90° ahead of the magnetizing
current space vector, iM- This is so since iM is what causes all the airgap flux that links the
stator and induces e s . For a given frequency, the amplitude of this voltage, e s , is proportional
to the current iM-

84 SYNCHRONOUS MACHINES AND DRIVES

1 m\

jx

M

Fig. 7.3 Stator equivalent circuit for a synchronous machine

Fig. 7.4 Phasor diagram of an synchronous machine

• A permanent magnet machine can be considered equivalent to that with a winding, carrying a
Direct Current, if, that is constant and cannot be controlled.

There are two modes of operation of a synchronous machine, that we'll study:

7.3 OPERATION OF THE MACHINE CONNECTED TO A BUS OF CONSTANT
VOLTAGE AND FREQUENCY

This is usually the case for large synchronous generators or motors. We can consider any bus as one
of constant voltage, by making a few modifications to the equivalent circuit as shown in figure 7.5.

OPERATION OF THE MACHINE CONNECTED TO A BUS OF CONSTANT VOLTAGE AND FREQUENCY 85

(a) Original Equivalent Circuit (b) The same circuit with Thevenin (c) Bqack to the Norton Equivalent cir-

with non-zero bus impedance Equivalent of the Synchronous Ma- cuit resulting in a modified Equivalent

chine Circuit

Fig. 7.5 Accounting for system impedance in the model of a Synchronous Machine

Synchronous machines are very efficient, and most of the time we can neglect the stator resistance.
All power then is converted to mechanical power and:

P = 3V S I S cos = Tcu s - (7.5)

P
P = -3V s I F cosf3 (7.6)

I M = Is + If (7.7)

V s = jX m I m (7.8)

In this operation V s and uj s (and therefore speed) remain constant. The only input variables are the
torque, T, which affects output power, P out = Tuu s -, and the field current, if, which is proportional
to If ; the magnetizing current Im is constant, since it is tied to the voltage V s .

Let us assume that the machine is operated so that the power to it varies while the frequency and
field current remain constant. Since this is a synchronous machine, the speed will not vary with the
load. From equation 7.6 we can see that the power, and therefore the torque, varies sinusoidally
with the angle (3. Remember that (3 is the angle between the axis of the rotor winding, and the
stator voltage space vector. Since this voltage space vector is 90° ahead of the space vector of the
magnetizing current, (3 — 90° is the angle between the rotor axis and the magnetizing current space
vector (same as the airgap flux). When there is no torque this angle is 0, i.e. the rotor rotates aligned
with the flux, but when external torque is applied to the rotor in the direction of rotation the rotor
will accelerate. As it accelerates (with the flux rotating at constant speed) the flux falls behind the
rotor, and negative torque is developed, making the rotor slow down and rotate again at synchronous
speed, but now ahead of the flux.

Similarly, when load torque is applied to the rotor, the rotor decelerates; as it does so, the angle /3
decreases beyond —90°, i.e. the rotor falls behind the flux. Positive torque is developed that brings
the rotor back to synchronous speed, but now rotating behind the stator flux.

In both cases when the load torque on a motor or the torque of the prime mover in a generator
increases beyond a maximum, corresponding to cos /3 = ±1, the machine cannot develop adequate
torque and it loses synchronization.

Let us discuss now the effect of varying the field current while keeping the power constant. From
equation 7.6, when power and voltage are kept constant, the product Ip cos /3 remains constant as
well. But this product is the projection of If on the horizontal axis. This means that as the field
current changes while power stays the same, the tip of If travels on a vertical line, as shown in figure
7.7a. Similarly, equation 7.5 means that at the same time the tip of I s travels on another vertical line,
also shown in figure 7.7a.

86 SYNCHRONOUS MACHINES AND DRIVES

Fig. 7.6 Torque and angle j3 in a synchronous motor

(a) Varying the field current in a Synchronous motor (b) Varying the field current in a Synchronous Gener-

under constant Power ator under constant Power

Fig. 7.7

It is clear from figure 7.7a that once the field current has exceeded a value specific to the power
level, the power factor becomes leading and the machine produces reactive power. This is different
from the operation of an induction machine, which always absorbs reactive power.

OPERATION OF THE MACHINE CONNECTED TO A BUS OF CONSTANT VOLTAGE AND FREQUENCY 87

When the machine operates as a generator, the input power is negative. Figure 7.7b shows this
operation for both leading and lagging load power factor. Here the angle between stator voltage
and stator current defined in the direction shown in the equivalent circuit, is outside the range

-90° < < 90°.

7.3.1 Example

A 3-phase Y -connected synchronous machine is fed from a 2300F, 60Hz. The ratio of the AC stator
equivalent current to the rotor DC is Ip /if = 1.8 The magnetizing inductance of the machine is
200mH.

• The machine is operated as a motor and is absorbing HOkW at 0.89 p.f leading. Calculate
the required field current and the load angle. Draw the corresponding phasor diagrams.

Using figure 7.8:

Fig. 7.8

X M = 2tt60 • 0.2 = 75.4ft
V. = ^ = 1328F

11Q.1Q3/3 Z27 , i0a

1328 • 0.89

from the stator voltage we can calculate 1m and from it Ip.

i M = ig = um^A

l F = I M - I s = 42 Z " 131 °A
=>i f =^- =23.4,4

J 1 Q

• Repeat for operation as generator at HOkW, 0.82 pf leading.
Using figure 7.9:

88 SYNCHRONOUS MACHINES AND DRIVES

-145

\ 35°

*►

I

_

^ — — ^Z3;56

V
s

7^

^

F

' ^ ^

M
Fig. 7.9

l g = 33.7 Z35 A => l s = 33.7 Z - 145 °A

I F = I M -Is = 27.66 Z3 - 56 °A

=>i f = 15.37 A

• What is the maximum power the machine above can produce (or absorb) when operating as a
generator and at the field current just calculated?

We know that absorbed and produced power is:

P = -3V s I F cos(3

for if = 15.37A we have I F = 27.66A, and P becomes maximum for (3 = 0, hence:

P = 3 • 1328 • 27.66 = 110.2kW

• If the terminal voltage remains at 23001^, 60 Hz, what is the minimal field current required to
maintain operation as a motor with load 70kW?

Again here:

P = 3 • V 3 If cos/3 = 3 • 1328 • I F = 70 • 10 3 W

=>I F = 17. ,57 A

7.4 OPERATION FROM A SOURCE OF VARIABLE FREQUENCY AND VOLTAGE

This operation requires that our synchronous machine is supplied by an inverter. The operation now
is entirely different than before. We no longer have an infinite bus, but rather whatever stator voltage
or current and frequency we desire. Moreover, with a space-vector controlled inverter, the phase
of this voltage or current can be arbitrarily set at any instant i.e. we can define the stator current

OPERATION FROM A SOURCE OF VARIABLE FREQUENCY AND VOLTAGE

89

or voltage space vector, and obtain it at will. The considerations for the motor operation are also
different:

• There is no concern for absorbing or supplying reactive power. Instead, there is a limit on the
total stator current, determined by thermal considerations.

• There is a limit to the maximum voltage the source can supply, which leads to modifications
of the machine mode of operation at high speeds.

Operation from source of variable frequency and voltage is most common for Permanent Magnet
Machines, where the value of |Ip| is constant.

In simple terms, when the machine is starting as a motor the frequency applied should be zero, but
the voltage space vector should be of such angle with respect to the rotor that torque is developed, as
discussed in the previous section. As torque develops, the machine accelerates, and the applied stator
currents have to create a rotating space vector leading the rotor flux. Voltage and frequency have to
be increased, so that this torque is maintained. It is important therefore to monitor the position of the
rotor in order to determine the location of the stator current or voltage space vector.

Two possible control techniques are implemented: either voltage control, where the stator voltage
space vector is determined and applied, or current control, where he stator current space vector is
applied.

For a fixed stator voltage and power (and torque) level, the stator losses are minimal when the
stator voltage and current are in phase. Figure 7.10 shows this condition.

Fig. 7.10 Operation of a synchronous PM drive at constant voltage and Frequency

Notice that as the power changes with the voltage constant two things happen:

1 . The voltage space vector varies in amplitude and the magnetizing current changes with it.

2. The amplitude of If stays constant, but its angle with respect to the voltage changes.

From the developed torque and speed we can calculate the frequency, the values of Im and Is,
and the angle between the stator voltage space vector and the rotor, since

3P V T - 3P T T T

- — V s l s — —LmJ-MJ-s

Zuj s z
I 2 M+Is

(7.9)
(7.10)

and Ip is a constant in PM machines.

90

SYNCHRONOUS MACHINES AND DRIVES

More common though is the case when the stator voltage is not constant. Here we monitor the
position of the rotor and since the rotor flux and rotor space current are attached to it, we are actually
monitoring the position of Ip . To make matters simple we use this current rather than the stator
voltage as reference, as shown in figure 7.11.

i's X M

Fig. 7.11 Operation of a synchronous PM drive below base speed

Although previous formulae for power and torque are still true they are not as useful. We create
new formulae that have the stator current I s and magnetizing current Im as variables. We also use
the angle 7, between 1^ and I s , since we can control it. Starting from what we already know:

P g = Sft[V s I F ] = 5R[JX m (If + I 8 )If]
= 5R[JX m IfI f ] + 5R[jX M IsI F ]

X M 3[I s Ii

XmIsI

m1 s 1fsiii7

(7.11)

(7.12)
(7.13)

For a given torque minimum losses require minimum value of the stator current. To minimize the
value of I s with constant power and Ip we choose 7 = 90° and arrive at:

T = 3

P_Pg_

'2 a;,,

Pg = X M % [I S I F ] = X M I S I F

= 3-L M ^ [I s Ip] = 3-L M I s l

(7.14)

(7.15)

which means that for constant power the projection of the stator current on an axis perpendicular to
Im is constant.

As the rotor speed increases, even if Im stays constant, the stator voltage V s = uo s L m I s increases.
At some speed ou s b, the required voltage exceeds the maximum the power source can provide. We
call this speed base speed; To increase the speed beyond it we no longer keep 7 = 90°. On the other
hand at that speed we know that the voltage has reached its upper limit V s = V s ^ max , therefore the
value of Im = V s ^max/^M is known. In this case, equations 7.9 and 7.10 become:

T =

T 2

J3p

2u s

V q I q cos 9

3p

L m ImIsCos9

2ImI s sin#

(7.16)

(7.17)

OPERATION FROM A SOURCE OF VARIABLE FREQUENCY AND VOLTAGE 91
J's1 X M M Y

Jl F A .

i' s2 X M

Fig. 7. 12 Field weakening of a PM AC motor. The two diagrams at are at the same frequency, but the second
one has 7 > 90° and lower V s

Figure 7.12 shows such an operation with the variables having the subscript 1. Note that we
calculate torque from power:

P = 3X M /5^Fsin7
T = —5 = 3? L M Z [I S I* ] = 3 P -L M I S I F sin 7

UJ s Z Z Z

(7.18)
(7.19)

7.4.1 Example

A 3-phase, four pole, Y connected permanent magnet synchronous machine is rated 400 V, 50Hz,
bOkVA. Its magnetizing inductance is 2.5mH and its equivalent field source current is 3 10 A. We
can neglect stator resistance.

• The machine is operated as a generator at rated frequency. Determine the maximum and
minimum values of the stator phase voltage as the load current is varied from zero to rated
value at unity power factor.

The rated phase voltage is V s = 400 /a/3 = 23 IF and the rated stator current is I s =
50 • 10 3 /3 • 231 = 72.2 A. With no load and at rated frequency the phase voltage is:

V s = co s L M I F = 2tt50 • 2.5 • 10" 3 • 310 = 243.51/

If the motor is operated at unity power factor, the stator current is collinear with the stator
voltage, as in figure 7.13.

From the current triangle:

i2 m = i2 f~ l2 s ^Im = V310 2 - 72.7 2 = 301.5A

and the stator voltage is:

V s = uj s L m Im = 236.8V

• The machine is now operated as a variable speed drive motor from a variable voltage, variable
frequency source. What should be the voltage and frequency in order to provide torque of
300Nm at 600rpm, if again we have unity power factor?

92 SYNCHRONOUS MACHINES AND DRIVES

is

The machine has four poles, so

Fig. 7.13

U 2lT

uo s = -600— = 20Hz
2 60

Torque can be expressed as a function of input power:

T = Z P -—VJ s pf = 3^(u s L M I M )I s = ^-L M I M I S = 300iVm

2 uj s 2uj s 2

In addition to this equation we have from the current triangle for unity pf:

i 2 F = i 2 M + i 2 s =i\o 2

These two equations, solved together will give

I M = 303A
I M = 66

I s = 66A or
L = 303A

which leads to phase voltage and torque:

v s =

p —

lo s L m I m = 95.11/

2

-u s T = 18.84/cW

P

7.4.2 Example

A 2-pole, Y -connected, 3-phase Permanent Magnet synchronous generator is rated 230F (l-l)
lOkVA, 400Hz. Its magnetizing inductance is 0.6mH. First a test is performed: The rotor is
externally driven at rated speed with the stator open circuited and the line-line voltage is measured
at 240K

Based on the result of this test determine the stator voltage and power angle when the stator
current, voltage and frequency are rated and the power factor of the load is 0.9 lagging.

OPERATION FROM A SOURCE OF VARIABLE FREQUENCY AND VOLTAGE 93

From the test:

X M = uo s L M = 2tt400 • 0.6 • 10
240

71*

K

IImXmI

~ 3 = 1.5081]
I M = 91-9-4

If — Im — 9 1.9 A and it is constant
Now that we found Ip, to the problem: At the operating point

J. = * = ^ = 25.102.4
y/Wu 73 • 230

pf = 0.9 => = -25.84°
from the geometry of the current triangle:

L M

I 2 M + I 2 s -2I M I s cos(^+e)=l\
f 25.1 2 + 2 • 25.1 • I M • 0.436 = 91.9

LM

100A

Fig. 7.14 Phasor diagram for this example

7.4.3 Example

A permanent magnet, Y connected, three-phase, 2-pole motor has If = 40 and Xm = 0.9D at
100Hz.

1. If it is absorbing P = 1 .5kW at 100Hz with minimum stator current I s , calculate this current,
the angle between I s and Ip, the speed, the stator voltage (line -neutral) and the power factor.

The minimum current I s will exist when 7 = Z(I S , Ip) = 90°. Then:

1500

P — 3XmI s If

3 • 0.9 • 40

13.89A

I M = If + I s = 40 + 13.89 Z90 ° = 42.34 Zl9 - 15 °A
=> V s = JWsLmIm = 38.12 Zl09 - 15 V

94

SYNCHRONOUS MACHINES AND DRIVES

the power factor is:

pf = cos(109.14° - 90°) = 0.946lagging

2. It is desired to increase the motor speed to 6900rpm while keeping power the same, P =
1.5kW, but the supplied voltage has reached its upper limit ofV s = 38.12V. Now the motor
absorbs the same power at at the voltage calculated in the previous question, but at frequency
llbHz; This can be accomplished by having stator current no longer at a minimum value and
7 7^ 90°. Calculate again the angle between I s and If, the speed, and the power factor.

-3V s If cos j3 => cos j3

1500

LM

>/3 =

3-38.12-40
109.14° => V,

-0.32

38.12 Zl09 - 14 °A

38.12

Zl09.15°

M

7 -ii5n Q
J ioo u -^

"-M

36.83 Zl9 - 15 °A

13.16 Zll3 - 18 °A

the power factor is now

pf = cos(109.14° - 113.18°) = 0.997leading

Casel

Case 2

Fig. 7. 15 Phasor diagram for this example

7.5 CONTROLLERS FOR PMAC MACHINES

Figure 7.16 shows a typical controller for an AC Machine. It requires a DC power supply, usually a
rectifier fed from an AC source, an inverter and a controller.
Figure 7.17 shows in a slightly higher detail the controller

BRUSHLESS DC MACHINES 95

3-

phase

AC

Rectifier

Inverter

Speed or Torque
Command

Voltage or

Current
Command

Controller

Fig. 7.16 Generic Controller for a PMAC Machine

PMAC
Motor

~~ PI \J* Calculate
Controller I

Calculate

sa' sb' sc

-► Inverter

Optical
encoder

4^h

Position of L

►()

+ f Ro tor position
1/s

| Rotor speed

Fig. 7.17 Field Oriented controller for a PMAC Machine. The calculations for I s are based on equation 7.19,
and the calculation of i* a , i* b , i* sc are calculated from the space vector I s from equations 5.3

7.6 BRUSHLESS DC MACHINES

While it would be difficult to find the difference between a PM AC machine described above and
a brushless DC machine by just looking at them, the concept of operation is quite different as is
the analysis. The windings in the stator in a brushless DC machine are not sinusoidally distributed
but instead they are concentrated, each occupying one third of the pole pitch. The flux density on
the magnet surface and in the airgap is also not sinusoidally distributed over the magnet but almost
uniform in the air gap.

As the stator currents interact with the flux coming from the magnet torque is developed. It should
be clear that for the same direction of flux, currents in opposite directions result in opposite forces,
and therefore in reduction of total torque. This in turn makes it necessary that all the current in the
stator above the rotor is in the same direction. To accomplish this the following are needed:

• Sensors on the stator that sense the direction of the flux coming from the rotor,

96

SYNCHRONOUS MACHINES AND DRIVES

• A fast supply that will provide currents to the appropriate stator windings as determined by
the flux direction.

• A way to control these currents, e.g. through Pulse Width Modulation

• A controller with inputs the desired speed, the flux direction in the stator and the stator currents,
and outputs the desired currents in the stator

Figures 7.18 and 7.19 show the rotor positions, the stator currents and the switches of the supply
inverter for two rotor positions.

— 1

r

\ 1

I

(a) Switch positions

(b) Machine cross section

Fig. 7. 18 Energizing the windings in a brushless DC motor

\

(a) Switch positions

(b) Machine cross section

Fig. 7. 19 Energizing the windings in a brushless DC motor, a little later

The formulae that describe the operation of the system are quite simple. The developed torque is
proportional to the stator currents:

T = k-I s (7.20)

At the same time, the rotating flux induces a voltage in the energized windings:

E = k-u;

(7.21)

BRUSHLESS DC MACHINES 97
Finally the terminal voltage differs from the induced voltage by a resistive voltage drop:

Vterm = E + I a R (7.22)

These equations are similar to those of a DC motor 4.4 - 4.6. This is the reason that although this
machine is entirely different from a DC motor, it is called brushless DC motor.

8

Line Controlled Rectifiers

The idea here is to draw power from a 1 -phase or 3-phase system to provide with DC a load. The
characteristics of the systems here are among others, that the devices used will turn themselves off
(commutate) and that the systems draw reactive power from the loads.

8.1 1- AND 3-PHASE CIRCUITS WITH DIODES

If the source is 1 -phase, a diode is used and the load purely resistive, as shown in figure 8.1 things
are simple. When the source voltage is positive, the current flows through the diode and the voltage
of the source equals the voltage of the load. If the load includes an inductance and a source (e.g. a
battery we want to charge), as in figure 8.2, the diode will continue to conduct even when the load
voltage becomes negative as long as the current is maintained.

u dtode i

+ -0H ^

Fig. 8. 1 Simple circuit with Diode and resistive Load

99

100

LINE CONTROLLED RECTIFIERS

v d\o6e v^

(a)

(d)

Fig. 8.2 Smple Circuit with Diode and inductive load with voltage source

8.2 ONE -PHASE FULL WAVE RECTIFIER

More common is a single phase diode bridge rectifier 8.3. The load can be modelled with one of
two extremes: either as a constant current source, representing the case of a large inductance that
keeps the current through it almost constant, or as a resistor, representing the case of minimum line
inductance. We'll study the first case with AC and DC side current and voltage waveforms shown in
figure 8.4.

If we analyze these waveforms, the output voltage will have a DC component V do :

V do = -V2V s ~0.9V s

(8.1)

where V s is the RMS value of the input AC voltage. On the other hand the RMS value of the output
voltage will be

V s = V d (8.2)

containing components of higher frequency.

Similarly, on the AC side the current is not sinusoidal, but rather it changes abruptly between I d
and -I d .

I 8l = -y/2I d = 0.9I d (8.3)

ONE -PHASE FULL WAVE RECTIFIER 101

Model of the
utility supply

111-.

Q^r v d

Fig. 8.3 One-phase full wave rectifier

* =

Fig. 8.4 Waveforms for a one-phase full wave rectifier with inductive load

and again the RMS values are the same

Id = I,

(8.4)

Giving a total harmonic distortion

TUB = ^ P % Isl ^ 48.43%

(8.5)

It is important to notice that if the source has some inductance (and it usually does) commutation
will be delayed after the voltage has reached zero, until the current has dropped to zero as shown
in figure 8.5. This will lead to a decrease of the output DC voltage below what is expected from
formula 8.1.

102 LINE CONTROLLED RECTIFIERS

(a)

(b)

(c)

(d)

D x A D 3 A

F/flf. 5.5 One-phase full wave rectifier with inductive load and source inductance

8.3 THREE-PHASE DIODE RECTIFIERS

The circuit of figure 8.3 can be modified to handle three phases, without using 12 but rather 6 diodes,
as shown in figure 8.6. Figure 8.7 shows the AC side currents and DC side voltage for the case of
high load inductance. Similar analysis as before shows that on the DC side the voltage is:

V do = -V2V LL = 1.35Vi L

TV

From figure 8.6 it is obvious that on the AC side the rms current, I s is

2 / d = 0.816/ d

while the fundamental current, i.e. the current at power frequency is:

7 8 i = -V6I d = 0.78I d

(8.6)

(8.7)

(8.8)

Again, inductance on the AC side will delay commutation, causing a voltage loss, i.e. the DC
voltage will be less than that predicted by equation 8.6.

CONTROLLED RECTIFIERS WITH THYRISTORS 103

S^4 £*>6 £#2

Fig. 8.6 Three-phase full- wave rectifier with diodes

(c)

K60»- — 120° -

-D 6 H

-D 6 H

-£3 H

0-

h Z) 2 *>

i c

1—^2-^

~D 5 ^\

^£5 —

Fig. 8.7 Waveforms of a three-phase full- wave rectifier with diodes and inductive load

8.4 CONTROLLED RECTIFIERS WITH THYRISTORS

Thyristors give us the ability to vary the DC voltage. Remember that to make a thyristor start
conducting, the thyristor has to be forward biased and a gate pulse provided to its gate. Also, to turn

104 LINE CONTROLLED RECTIFIERS

off a thyristor the current through it has to reverse direction for a short period of time, t rr , and return
to zero.

8.5 ONE PHASE CONTROLLED RECTIFIERS

Figure 8.8 shows the same 1 -phase bridge we have already studied, now with thyristors instead of
diodes, and figure 8.9 shows the output voltage and input current waveforms. In this figure a is the
delay angle, corresponding to the time we delay triggering the thyristors after they became forward
biased. Thyristors 1 and 2 are triggered together and of course so are 3 and 4. Each pair of thyristors
is turned off immediately (or shortly) after the other pair is turned on by gating. Analysis similar to

"4 r 3 "i

riS t 3

T*

/N—

»d

cl

GK

Fig. 8.8 One-phase full wave converter with Thyristors

that for diode circuits will give:

V n

do

-v2V s cos a = 0.9Vs cos a

(8.9)

and the relation for the currents is the same

I sl = -V2I d = 0.9I d

(8.10)

We should notice in figure 8.9 that the current waveform on the AC side is offset i time with respect
to the corresponding voltage by the same angle a, hence so is the fundamental of the current, leading
to a lagging power factor.

On the DC side, only the DC component of the voltage carries power, since there is no harmonic
content in the current. On the AC side the power is carried only by the fundamental, since there are
no harmonics in the voltage.

P = V s I s iCosa = V d I d

(8.11)

8.5.1 Inverter Mode

If somehow the current on the DC side is sustained even if the voltage reverses polarity, then power
will be transferred from the DC to the AC side. The voltage on the D side can reverse polarity when

ONE PHASE CONTROLLED RECTIFIERS 105

(a)a =

3,4^4*-

(b) a = finite

Fig. 8.9 Waveforms of One-phase full wave converter with Thyristors

the delay angle exceeds 90°, as long as the current is maintained. This can only happen when the
load voltage is as shown in figure 8.10, e.g. a battery.

H

r~ T- — ^

1 "S 3 ~A (i>t hy ) 3

v d

Fig. 8.10 Operation of a one-phase controlled Converter as an inverter

106 LINE CONTROLLED RECTIFIERS

8.6 THREE-PHASE CONTROLLED CONVERTERS

[70-

a

-o-

l b

— ^)— O— ' >— / N-

«fc

tilt

tf

F/flf. 5. 7 7 Schematic of a three-phase Full- Wave Converter

fd

(d)

Fig. 8. 12 Waveforms of a Three-phase Full-Wave Converter

As with diodes, only six thyristors are needed to accommodate three phases. Figure 8.11 shows
the schematic of the system, and figure 8.12 shows the output voltage waveform. The delay angle a
is again measured from the point that a thyristor becomes forward biased, but in this case the point
is at the intersection of the voltage waveforms of two different phases. The voltage on the DC side

*NOTES 107

is then:

3

Vdo = -V2V LL cosa = 1.35V LL cosa (8.12)

7T

while the power for both the AC and the DC side is

P = V d I do = 1.35VuIdcosa = VsVuI 8l cos a (8.13)

I sl = 0.7SI d (8.14)

Again if the delay angle a is extended beyond 90°, the converter transfers power from the DC
side to the AC side, becoming an inverter. We should keep in mind, though, that even in this case
the converter is drawing reactive power from the AC side.

8.7 *NOTES

1. For both 1 -phase and 3 -phase controlled rectifier delay in a creates a phase displacement of
the phase current with respect to the phase voltage, equal to a. The cosine of this angle is the
power factor of the first harmonic.

2. For both motor and generator modes the controlled rectifier absorbs reactive power from the
three-phase AC system, although it can either absorb or produce real power. It also needs the
power line to commutate the thyristors. This means that inverter operation is possible only
when the rectifier is connected to a power line.

3. When a DC motor or a battery is connected to the terminals of a controlled rectifier and a
becomes greater then 90°, the terminal DC voltage changes polarity, but the direction of the
current stays the same. This means that in order for the rectifier to draw power from battery
or a motor that operates as a generator turning in the same direction, the terminals haver to be
switched.

9

Inverters

Although the AC-to-DC converters we have already studied can transfer power from the DC side to
the AC system, they require the presence of such an AC system in order to commutate the thyristors
and provide the required reactive power. In this chapter we'll study a similar system using devices
that we can turn both on and off, like GTOs, BJTs IGBTs and MOSFETs, which allows the transfer
of power from the DC source to any AC load. Figure 9.1 shows a typical application of a complete
system, where the supply power of constant voltage and frequency is rectified, filtered and then
inverted to provide an output of desired voltage and frequency.

We'll study first the operation of a single phase inverter and then we'll expand to three-phases.

9.1 1 -PHASE INVERTER

Figure 9.2 shows the operation of on leg of the inverter regardless of the number of phases. To
illustrate the point better, the input DC voltage is divided into two equal parts. When the upper
switch Ta+ is closed, the output voltage Vao W1 U be |Vd, and when the lower switch Ta- is closed,
it will be —\Vd. Deciding which switch to close in order to obtain a certain waveform will be
determined by the PWM comparison shown in figure 9.3. We define as the frequency modulation
index the ratio of the frequencies of the carrier (triangular wave) to the control signal:

h

and as amplitude modulation index:

v control /rv ^x

m a = — (9.2)

V+ri

1 . The output voltage in figure 9.3 at first look does not resemble the expected waveform (i.e. the
control signal). Its fundamental, though, does, and one can filter out the higher harmonics.

109

110 INVERTERS

+

60 Hz

i

J

I

/ ac

ac

t motor

Diode-rectifier Filter Switch -mode
capacitor inverter

Fig. 9. 1 Typical variable voltage and frequency system supplied from a power system

+

Yd 4-

2

V l D * +

il

V AN

Fig. 9.2 One leg of an inverter

f control v tri

!! ! l~K*>

A

v Ao, fundamental = ( v A<dl

Tf 1 "'

2

±

C "controK </ tri \ _J I
\T A _:on,T A + :attJ ^ -C

t =

"f ^control > i> tr j "I
\T A+ : on, 7^ off/

(b)

Fig. 9.3 PWM scheme to determine which switch should be closed

V A Qo

A

T A -

1

*>A~

THREE-PHASE INVERTERS 111

* 1* j 1 n

A

- u tf = ^o ~ *>#o

J

D B -

Fig. 9.4 One-phase full wave inverter

2. The switches in the inverter can conduct only in one direction. Inductive loads, though, require
the current to continue to flow after a switch has been turned off. Allowing this current to flow
is the job of the antiparallel diodes.

A full bridge inverter is shown in figure 9.4. It has four controlled switches, each with an
antiparallel diode, and diagonally placed switches operate together. The output voltage will oscillate
between +Vd and —Va and the amplitude of the fundamental of the output voltage will be a linear
function of the amplitude index V Q = mVd as long as m a < 1. Then the rms value of the output
voltage will be:

Ki = ^y=0.353m a ^

(9.3)

When m a increases beyond 1, the output voltage increases also but not linearly with it, and can reach
peak value of -Vd when the reference signal becomes infinite and the output a square wave. Its RMS
value, then will be:

7T Z

Equating the power of the DC side with that of the AC side

P = V d Ido = V ol I olP f
Hence for normal PWM

and for square wave

I do = 0.353m a I o ipf
I d o = 0.45/ O ip/

(9.4)

(9.5)

(9.6)
(9.7)

9.2 THREE-PHASE INVERTERS

For three-phase loads, it makes more sense to use a three-phase inverter, rather than three one-phase
inverters. Figure 9.5 shows a schematic of this system:

The basic PWM scheme for a three-phase inverter has one common carrier and three separate
control waveforms. If the waveforms we want to achieve are sinusoidal and the frequency modulation
index rrif is low, we use a synchronized carrier signal with rrif an integer and multiple of 3.

112 INVERTERS

l d

== r y W r y L B+ T y L c+

Yi

H v.

^- v

r ? i t

*,- y

i^c

N

oA f)B 6C

F/g. 9.5 Three-phase, full-wave inverter

^control, B "control, C

T

' _

TinnniiP"

I

y >4fl = <MN ~ ^N

..Fundamental u LL1

T

w

Fig. 9.6 Three-phase Pulse Width Modulation

THREE-PHASE INVERTERS 1 13

T *y Ad a+ J kD B +J adc + T *+ Vd

v d i

T A -

Fig. 9.7 6- step operation of a PWM inverter

As long as m a is less than 1, the rms value of the fundamental of the output voltage is a linear
function of it:

V L li = ^ m aVd ^ 0M2m a V d (9.8)

On the other hand, when the control voltage becomes infinite, the rms value of the fundamental
of the output voltage becomes:

V2tt 2

(9.9)

In this case the output voltage becomes rectangular and the operation is called 6- step operation, as
shown in figure 9.7b.

Equating the power on the DC and AC sides we obtain: Equating the power of the DC side with
that of the AC side

P = V d I d0 = V3V lll I olP f (9.10)

Hence for normal PWM
and for square wave

I do = 1.06m a I ol pf

i- do

1.35/ ip/

(9.11)

(9.12)

Finally, there other ways to control the operation of an inverter. If it is not the output voltage
waveform we want to control, but rather the current, we can either impose a fast controller on the
voltage waveform, driven by the error in between the current signal and reference, or we can apply
a hysteresis band controller, shown for one leg of the inverter in figure 9.8

114 INVERTERS

v d dtz

/t a+

AA4+

(T A .

i^-

Comparator

tolerance

band

Jd T<^

*A

ih

Reference current i A
Actual current i A

flown nn

T A - '• on T A + '■ on

(a)

Switch -

Mode

Inverter

,-Al

-A
-B
•C

(b)

Fig. 9.8 Current control with hysteresis band

THREE-PHASE INVERTERS

115

Notes

• With a sine-triangle PWM the harmonics of the output voltage s are of frequency around nf s ,
where n is an integer and f s is the frequency of the carrier (triangle) waveform. The higher
this frequency is the easier to filter out these harmonics. On the other hand, increasing the
switching frequency increases proportionally the switching losses. For 6-step operation of a
3-phase inverter the harmonics are even except the triplen ones, i.e. they are of order 5, 7, 11,
13, 17 etc.

• When the load of an inverter is inductive the current in each phase remains positive after the
voltage in that phase became negative, i.e. after the top switch has been turned off. The
current then flows through the antiparallel diode of the bottom switch, returning power to the
DC link. The same happens of course when the bottom switch is turned off and the current
flows through the antiparallel diode of the top switch.

9.2.1 Example

A 3-phase controlled rectifier is supplying a DC motor with k
fed from a 208V I — I source.

IV s and R = 1Q. The rectifier is

208V

Fig. 9.9 figure for 9.2.1

la Calculate the maximum no-load speed of the DC motor.
Without load the current is zero. Hence:

V = kou + IR = kuj

The maximum speed is then determined by the maximum DC voltage:

This maximum DC voltage is provided by the controlled rectifier for a = 0:

V max = 1.35V/Z = 281.8V

hence

lb The motor now is producing torque of 20Nm. What is the maximum seed the motor can
achieve ?

Now that there is load torque there is current:

T = kI=>I = 20Nm
V -IR 280.8 -20-1

Again

UJ

k

1

116

INVERTERS

lc For the case in lb calculate the total rms current the first harmonic and the power factor at
the AC side.

The fundamental of the AC current is

I si = 0.78I d = 15.6 A

Power factor is then 1.

Id The motor is now connected as a generator, with a counter torque of 20 Nm at 1500rpm.
What should be the delay angle and AC current?

For a DC generator

V = ku-IR = ku- -Rl • 1500 1 = 137.08V

k 60 1

Since this is a generator this voltage is negative for the inverter (see notes)

-137,08 = 1.35 • 208 cos a => cos a = -0.488 =>- a = 119.22°

9.2.2 Example

In the system below the AC source is constant. The load voltage is 150V(Z — I), 20Am 52Hz,
O.S5pf lagging. Calculate:

!08V

- u

/

Filter

6-step
inverter

a

Fig. 9.10 figure for 9.2.2

a The voltage on the DC side and the DC component of the current.
For 6-step inverter

Vh,i = 0.78V d => V d = 192V

r,r r r rr r r 1.35 • 150 • 20 • 0.85

P = y/3VuIipf = V d I d0 => /do = j<j2 = 23A

b Calculate the source side (208V AC) rms and fundamental current and power factor.
For a 3 -phase rectifier

V d = 1.35V// cos a => 192 = 1.35 • 208 • cos a => cos a = 0.685

I 8l = 0.78I d = 17.94A
pf = cos a = 0.685 lagging

10

DC-DC Conversion

We will study DC to DC converters operating under certain conditions. The use of such converters are
extensive in automotive applications, but also in cases where a DC voltage produced by rectification
is used to supply secondary loads. The conversion is often associated with stabilizing, i.e. the input
voltage is variable but the desired output voltage stays the same. The converse is also required, to
produce a variable DC from a fixed or variable source. The issues of selecting component parameters
and calculating the performance of the system will be addressed here. Since these converters are
switched mode systems, they are often referred to as choppers.

1 0.1 STEP-DOWN OR BUCK CONVERTERS

The basic circuit of this converter is shown in figure 10.1 connected first to a purely resistive load.
If we remove the low pass filter shown and the diode the output voltage v (t) is equal to the input
voltage Vd when the switch is closed and to zero when the switch is open, giving an average output
voltage V :

y-h

ton r

V d dt+ I Odt

J t on

^V d (10.1)

J- s

with t on /T s = D, the duty ratio.

The low pass filter attenuates the high frequencies (multiples of the switching frequency) and
leaves almost only the DC component. The energy stored in the filter inductor (or the load inductor)
has to be absorbed somewhere other than the switch, hence the diode, which conducts when the
switch is open.

We'll study this converter in the continuous mode of operation i.e. the current through the inductor
never becomes zero. As the switch opens and closes the circuit assumes one of the topologies of
figure 10.2.

117

1 18 DC-DC CONVERSION

+ 1 Low-pass

J filter

/ r "i 1 J^

v. + 1| r I .

- -f i **-frr-

I I

(a)

V d

"E

F/flf. 70. 7 Topology of the buck chopper

Fig. 10.2 Operation of the buck chopper

We'll use the fact that the average voltage across the inductor is zero. Assuming perfect filter, the
voltage across the inductor is Vd during t on and —V the remaining of the cycle. Hence:

/'

Jo

(V d - V )dt + / {-V )dt =

(V d - V o )t

V (T S - t on ) =
V d = T~ s

D

(10.2)
(10.3)
(10.4)

STEP-UP OR BOOST CONVERTER

119

A second consideration is that the input and output powers are the same, hence:

V d I d = V I
L = Vd = l_
la Vo D

(10.5)
(10.6)

Note that in discontinuous mode the output DC voltage is less that that given here, and the chopper
less easy to control.

At the boundary between continuous and discontinuous mode, the inductor current reaches zero
for one instant every cycle, as shown in figure 10.3a. Using this figure we can see that at this operating
point, the average inductor current is II = \ih- Further studying the geometry we obtain:

lL=\toniV d -V )

2L

(V d - V )

(10.7)

Since the average inductor current is the average output current (the average capacitor current is
obviously zero), equation 10.3 defines the minimum load current current that will sustain continuous
conduction.

Fig. 10.3 Operation of the buck Converter at the boundary of Continuous Conduction

Finally a consideration is the output voltage ripple. We assume that the ripple current is absorbed
by the capacitor, i.e. the voltage ripple is small. The ripple voltage is then due to the deviation from
the average of the inductor current as shown in figure 10.4. Under these conditions:

A^n

AQ = 1 1 AIl T s

C ~ L2 2 2

where AI L = Y±{\-D)T a

AV

Vo

L

1 T 2

_ s

8LC

(1-D)

Another way to look at this is to define the switching frequency f a
frequency of the filter, f c = 1/(2ttVLC):

AK

V 2 { } \U

(10.8)

(10.9)

(10.10)

1/T S and use the corner

(10.11)

1 0.2 STEP-UP OR BOOST CONVERTER

Here the output voltage is always higher than the input. The topology is shown in figure 10.5.

120 DC-DC CONVERSION

(Vd - »o)

(-O

Fig. 10.4 Analysis of the output voltage ripple of the buck Converter

*0

L

+ + v L -

-

i

= *<

+

F/flf. 70.5 Schematic Diagram of a Boost Converter

There are two different topologies, based on the condition of the switch, as shown in figure 10.6
Again, the way to calculate the relationship between input and output voltage we have to take the
average current of the inductor to be zero, and the output power equal to the input power hence:

r dt on + (Vd ~

- V )(T S - t on ) =

_ Vo_ 1

^ Vd 1-D

^^ = 1-D

id

(10.12)
(10.13)

(10.14)

STEP-UP OR BOOST CONVERTER 121

<Vd)

(V d ~ VJ

Fig. 10.6 Two Circuit Topologies of the boost Converter

To determine the values of inductance and capacitance we will study the boundary of continuous
conduction like before and the output voltage ripple.

h = lLB

I V ~ Constant

!><

-'«.«•

Wo

/VcB.nm* =

0.074 T S V^

( . . . . I I

_l

! ^V

I ^

I

S*IB

0.25 | 0.5

0.75 1.0

F/(/. 70.7 The boundary between Continuous and Discontinuous Conduction of a Boost Converter

At the boundary of the continuous conduction, as shown in figure 10.7, the geometry of the current
waveform will give:

T S V _ _ 2

2L

-D(l-D) 2

(10.15)

The output current has to exceed this value for continuous conduction. Looking at the geometry
of figure 10.8 and following an analysis similar to that of a buck converter we find that:

AV DT S

V RC

(10.16)

It is important to note that the operation of a boost converter depends on parasitic components,
especially for duty cycle approaching unity. These components will limit the output voltage to levels
well below those given by the formula 10.13.

122 DC-DC CONVERSION

Fig. 10.8 Calculating the output voltage ripple for a boost inverter

1 0.3 BUCK-BOOST CONVERTER

This converter, the topology of which is shown in figure 10.9, can provide output voltage that can be
lower or higher than that of the input.

*rf

•/

vt

R

Fig. 10.9 Basic buck-boost converter

Again the operation of the converter can be analyzed using the two topologies resulting from
operation of the switch, shown in figure 10.10.

By equating the integral of the inductor voltage to zero we can get:

V d DT a + (-V o )(l-D)T 8 =

Vo = D
^ V d 1-D

(10.17)
(10.18)

At the boundary between continuous and discontinuous conduction we can use figure 10.11 to
find that

2L

{1-Df

(10.19)

BUCK-BOOST CONVERTER 123

w*

11

T
X

Fig. 10.10 Operation of a buck boost chopper

1.0

\

0.75

" N \. |^ = Constant

0.50

^ ^V T,V
\ ^\Jlb, max = IoB, max = ~7f~

\^ \^~IlbUlB, max

t 0.25

\ X

^oB^oB, max" \ X^

\ X

1 1 T.^-J* „n

0.25 0.50 0.75 1.0

(b)

Fig. 10.11 Operation of a buck boost chopper

The output voltage ripple, as calculated based on figure 10.12 is

AV

v

D-

RC

(10.20)

10.3.1 Example

The input of a step down converter varies from 301/ to AOV and the output voltage is to be constant
20V, with output power varying between 100W and 200 W. The switch is operating at 20kHz.
What is the inductor needed to keep the inductor current continuous ? What is then the filter capacitor
needed to keep the output ripple below 2%.

The duty cycle will vary between D\ = 20/30 = 0.667 and D 2 = 20/40 = 0.5. The load current
will range between I ol = 100/20 = 5A and I o2 = 200/20 = 10A

124 DC-DC CONVERSION

Fig. 10. 12 Calculating the output voltage ripple for a boost inverter

The minimum current needed to keep the inductor current continuous is

DT

°(V d -V )

since the constant is the output voltage V and the minimum load current has to be greater than
lo min> we'll express it as a function ofV and make it less or equal to bA!

f)A> I

DT V T

HV d -V ) = ^(l-D)

2L

2L

T s = 1 /20kHz, V = 20V and the max value is achieved for D = 0.5, leading to L m i n = 50fiH.
As about the ripple, the highest will occur at 1 — D = 0.5. Hence:

0.02 = ^-0.5

fc

2 * V 10 • 10 3
1

27TA/50 • 10-6C
=> C = 625/iF

f c = 900Hz

900

``` 